Chinese Remainder Theorem

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Modulo Arithmetic Remainder Theoryby fundoo bond - Friday, 28 September 2007, 03:18 AM

This extremely useful article for finding remainders has been contributed to TG by Fundoo Bond, anotherintelligent TGite. All the CAT 2007/2008 who find this article useful are requested to say thanks to Fundoo Bondfor his efforts and this wonderful compilation. If you have some interesting article, funda or helpful informationto share, please mail it to us and we will post it by your name- Total Gadha

The problems of finding the remainder are considered the most dreadful among the number theory problems.Fortunately, if we arm ourselves with some basic theorems, we will see that we can turn these mind-bogglingproblems into sitters and can ensure some easy marks.

I will be trying to share both the theory and the examples in order to make the concepts clearer. Again friends, do keep in mind that Iam not a genius to write these theorems on my own. I have learnt them from various sources and I am trying my best to explain you inmy own words.



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Re: Modulo Arithmetic Remainder Theoryby Rishi Kapoor - Friday, 28 September 2007, 05:06 AM

Thanks Sir...This is what I was searching for...and you have given it in the best time...

You are really TG.....Toooooooo Gooooood

with these Fundoo theorams, you have proved yourself BOND

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Re: Modulo Arithmetic Remainder Theoryby Mady Ahmed - Friday, 28 September 2007, 05:36 AM

wow,fabulous, fantastic,m ind blowing,excellent ...words are short of my dictionary to praise u thanx alot

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Re: Modulo Arithmetic Remainder Theoryby shyam Sundar - Friday, 28 September 2007, 08:46 AM

hey folks

how to do 40!%83

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Re: Modulo Arithmetic Remainder Theoryby AD P - Friday, 28 September 2007, 09:25 AM

corollary of the Wilson theorem which will be very useful when solving remainder problems.From Wilson theorem,we have (p-1)!%p = (p-1) or -1----(1)from (1) we can say (p-2)!%p = 1

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Hey Shyam SunderSolution to ur problem

40! mod 83We know tht--frm Wilson's Theorem(p-1)!%p = (p-1) or -1(iff p is prime)so, we can deduce82!%83=-1 or 82---------------(1)frm Wilson's Corollary(p-2)!%p = 1so, we can deduce here81!%83=1-----------------(2)Now 81!=81.80.79.-------.42.41!----------(3)from 2 and 3(81.80.79.--------.42.41!)%83=1-2*-3*-4*-5....-41*41! = 41!*41!%83 = (41!)^2%83 = 1which is possible only when 41! mod 83=1because there are even no of terms so negative sign got cancelled

Now we have 41!mod83=1(41.40!)%83=1(-42.40!)%83=1suppose 40!=x(-42*x)%83=1which is possible only when x becomes -2 or 81so ans is 81

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Re: Modulo Arithmetic Remainder Theoryby subhadeep das - Friday, 28 September 2007, 09:43 AM



how to solve (i) find the remainder of 55555 .... 93 times divided by 98 (ii) remainder of 2 ^1990 / 1990

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how to solve (ii) remainder of 2 ^1990 / 19901990=2.5.199from Euler's Theoremphi(2)=1phi(5)=4phi(199)=1982^4=1 mod 5--------(1)2^198=1 mod 199---------(2)LCM(4,198)=396so2^396=1 mod 19902^1980=1 mod 1990(because 1980 is a multiple of396)so now we remain with2^10 mod 19901024 mod 1990=1024ans is 1024

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Re: Modulo Arithmetic Remainder Theoryby lavika gupta - Friday, 28 September 2007, 09:51 AM

hi Mr. Fundoo Bond.

this is a masterpiece 2 learn the concepts of remainder theorem.i found the chinese remainder theorem a bit more abstruse 2 master,yetother theorems and their enunciation with examples proved really helpful 2 me.hope 2 see such great work again from u.

HATS OFF TO U

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Re: Modulo Arithmetic Remainder Theoryby Catapult!!! On the way - Friday, 28 September 2007, 09:58 AM

Hi Fundoo Bond!!!

Just one word..

"Superb"..

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(i) find the remainder of 55555 .... 93 times divided by 98 98=2.7.7It is clear that the given no gives remainder of 11.....11 93 timeswhen divided by 2so now question remains(5555....55 93 times) mod 4955555.......55 93 times=5x10^92+5x10^51+--------+5=5[10^0+10^1+--------+10^92]=5[(10^93-1)/9]=5[(11111111111111....11 92 times)]I think ans is 5plzz check it out

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Re: Modulo Arithmetic Remainder Theoryby jitendra havaldar - Friday, 28 September 2007, 10:23 AM

Hi TG,

really an unbelievable article to say the least ....

Continue your good work ...

Thanks,

Jitendra



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Re: Modulo Arithmetic Remainder Theoryby Tom Goel - Friday, 28 September 2007, 10:36 AM

hi fundoo,

Though we all know you are a fundoo and a bond.....by sharing this article to have shown altruistic character.Hats off to you.

Thanks

Amit

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Re: Modulo Arithmetic Remainder Theoryby Sanju P - Friday, 28 September 2007, 11:03 AM

Hi Fundoo,

Wonderful article and great compilation!

Thanks!

S P

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Re: Modulo Arithmetic Remainder Theoryby Software Engineer - Friday, 28 September 2007, 11:18 AM

FB

Marvelous Methods ! ! !

Thanks,SE

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Re: Modulo Arithmetic Remainder Theoryby purvi verma - Friday, 28 September 2007, 12:18 PM

fabulous is the word.....please post some more fundas re lated to numbers.

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Re: Modulo Arithmetic Remainder Theoryby ankit harish - Friday, 28 September 2007, 12:32 PM

Awesome Compilation.Thanks a lot bond

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Re: Modulo Arithmetic Remainder Theoryby fundoo bond - Friday, 28 September 2007, 12:40 PM

Hi TG n all TGites,

Thanks a lot for a ll the appericiation....Feeling great that my efforts are appericiated...And it gives a grt joy that the article was published

exactly as it was submitted ...

regards,fundoo

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Re: Modulo Arithmetic Remainder Theoryby Dagny Taggart - Friday, 28 September 2007, 01:20 PM

Kudos Fundoo

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Re: Modulo Arithmetic Remainder Theoryby Total Gadha - Friday, 28 September 2007, 01:55 PM

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Re: Modulo Arithmetic Remainder Theoryby SS VV - Friday, 28 September 2007, 03:01 PM

i'm sure u wont believe it ...but was just going to post a request to you ask ing you to expla in these very same theorems ....had seen themin one of your post wherein tg mentioned these are your favourite topics:>

secondly wanted to thank you for a ll your previous posts...u really took grt pains in expla ing all your methods in detail( remember the onein which i asked for digita l root and you typed up the detailed approach) and now this....wht can i say but 'rock on dude'

cheers!!

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Re: Modulo Arithmetic Remainder Theoryby imran khan - Friday, 28 September 2007, 03:58 PM

u guyz are jus fantabulous.............thanx a ton......god bless ur species............ .

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Re: Modulo Arithmetic Remainder Theoryby Manish Kashyap - Friday, 28 September 2007, 04:51 PM

Hi Fundoo,

Its really superb.Only one thing I can say "Fundoo".

Thanks and Regards,

Manish

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Re: Modulo Arithmetic Remainder Theoryby Shubhagata roy - Friday, 28 September 2007, 07:00 PM

bindassss............rem of 7^115 divided by 114??

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Re: Modulo Arithmetic Remainder Theoryby Abhra Chatterjee - Friday, 28 September 2007, 07:25 PM

Wonderful job Bond ..... P lease post some thing on Permutation and Probability.

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Re: Modulo Arithmetic Remainder Theoryby Aviator D - Friday, 28 September 2007, 07:29 PM

hi fb,

First, thank you for the post. It is very helpful.

In the ques. : how to solve (ii) remainder of 2 ^1990 / 19901990=2.5.199from Euler's Theoremphi(2)=1phi(5)=4phi(199)=1982^4=1 mod 5--------(1)2^198=1 mod 199---------(2)LCM(4,198)=396so2^396=1 mod 19902^1980=1 mod 1990(because 1980 is a multiple of396)so now we remain with2^10 mod 19901024 mod 1990=1024ans is 1024

how can we apply uler's theorem in this case? because according to ur post we can only use the theorem whenboth M(=2) and N(=1990) are coprimes, but this not the case here.



Kindly elaborate.

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Re: Modulo Arithmetic Remainder Theoryby Deep Agrawal - Friday, 28 September 2007, 07:55 PM

7^115 / 114.

Phi(114) = 66

hence 7^66 / 114 = 1

Remaining is 7^49 /114

=> ((7^3)^16 * 7) / 114 = 1^16 * 7 / 114 = 7 (Since, 7^3 = 343 => 343/114 = 1)

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Re: Modulo Arithmetic Remainder Theoryby rohit avasthi - Friday, 28 September 2007, 08:30 PM

hey,

thnx for this article...i hav been luk in for the totient style of solving method since last few months

the other methods were as fascinating...

......it was extremely helpful...

thnx

rohit

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Re: Modulo Arithmetic Remainder Theoryby rohit avasthi - Friday, 28 September 2007, 08:18 PM

hey,

thnx for this article...i hav been luk in for the totient style of solving method since last few months......it was extremely helpful...

thnx

rohit

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Re: Modulo Arithmetic Remainder Theoryby subhadeep das - Friday, 28 September 2007, 10:21 PM

@AD Pyah , both the answers that you got are right , but i'm having tough time understanding your soln for 2^1990/1990 , specia lly the part

phi(2)=1phi(5)=4phi(199)=1982^4=1 mod 5--------(1)2^198=1 mod 199---------(2)LCM(4,198)=396

please e laborate a little bit more

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Re: Modulo Arithmetic Remainder Theoryby AD P - Friday, 28 September 2007, 10:24 PM

Hi Aviator DI followed the Euler's TheoremEuler's Theorem is the generalization of Fermat's Little Theorem, applicable when ur dividend is composite in form.So we first find primefactors for the composite no.-----After getting prime factors, use Euler's totient to obtain phi(x) values. phi(x)just define no of no which are coprime to x &



Re: Modulo Arithmetic Remainder Theoryby AD P - Friday, 28 September 2007, 10:33 PM

Hi ShubhdeepWat is Euler's Theorem

Eulers Theorem:

IF x is any positive integer prime to N then x^Phi(N)=1(mod N)

N=a^p.b^q.c^r.(prime factors)

Phi(N)=N(1-1/a)*(1-1/b)*(1-1/c) and is known as Eulers Totient Function

Phi(N) defines no of numbers which are less thenN and coprime to N

If N in the above Eulers Theorem is prime,then Phi(N)=N(1-1/N)=

(N-1)

So if N is prime and x is coprime to N,then the Eulers Theorem becomes Fermats Theorem

As x^(N-1)=1 (mod N)

So in this question

we first break 1990 into prime factors because for Euler condition is that x & N are prime to each other i.e coprimes

1990=2.5.199

and separately apply Euler

hope so its clear now

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Re: Modulo Arithmetic Remainder Theoryby indira ghosh - Friday, 28 September 2007, 11:30 PM

thanks . i'll practice & try a ll this method. It was great . but i have problems in geometry so what i will do ?

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Re: Modulo Arithmetic Remainder Theoryby Aviator D - Saturday, 29 September 2007, 07:56 AM

@AD P - Thanks for the prompt response.

Hi Aviator DI followed the Euler's TheoremEuler's Theorem is the generalization of Fermat's Little Theorem, applicable when ur dividend is composite in form.So we first find primefactors for the composite no.-----After getting prime factors, use Euler's totient to obtain phi(x) values. phi(x)just define no of no which are coprime to x &



2^phi(5)=2^4=1 mod 52^phi(199)=2^198=1 mod 199

take LCM of powers i.e. 4 & 198 here

LCM(4,199)=396so we can say2^396=1 mod 1990---simplehope u understand

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Re: Modulo Arithmetic Remainder Theoryby mon haldi - Saturday, 29 September 2007, 11:11 AM

remainder of 5555......93times divided by 98

5555555....93 times= 5(111111....93times)= 5{11(10^91 + 10^89 + ...+10^3 + 10^1) +1}= 5{11*10(10^90 + 10^88 +....+10^2 + 1) + 1}......Eq(1)

now 98 can be written as = 100-2 or = 10^2 -2use remainder theorem method and put 10^2 = 2 in Eq(1)

= 5{110(2^45 + 2^44 + ... + 1) +1}= 5{110(2^46-1) +1} .....Eq(2)remainder of 2^46 when divided by 98 is 16 (do own your own but be careful remainder 2^10 / 98 is 44 )

= 5{110(16-1) + 1}= 5(12*15 +1)= 5(-16 +1)= 5 (-15)= -75= 23so remainder is 23let me know if this is the solution

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Re: Modulo Arithmetic Remainder Theoryby prafull kumar - Saturday, 29 September 2007, 04:37 PM

Hi Fundooooooooo, just want to say

Tooooooooooooooooooooooooooo Goooooooooooooooooooooooooooooooooooooooooood

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Re: Modulo Arithmetic Remainder Theoryby prabhjeet kaur - Saturday, 29 September 2007, 06:10 PM

helllooo sir ..

thankss for such a excellent concept but sir i cldnt be able to take out the print of ths page !! wat shld i do now ?

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Re: Modulo Arithmetic Remainder Theoryby harshal malaviya - Saturday, 29 September 2007, 11:18 PM

Thank you,sir,the article is amazing.i was searching for these concepts from a long time...god bless u guys, u guys r doin a damn gudjob...some fundas on algebra,would be of great help,especia lly on hw. to solve prob.on graphs and solving prob.with the help ofgraphs...thankin u .

harshal.

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Re: Modulo Arithmetic Remainder Theoryby AD P - Saturday, 29 September 2007, 11:28 PM

@mon Haldiur solution is correctgud solution

can u help me in following question

40! mdo 83

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Re: Modulo Arithmetic Remainder Theoryby Naveen Naveen - Sunday, 30 September 2007, 12:34 AM

This is really a great stuff...and this site is cool

hanks a lot...

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Re: Modulo Arithmetic Remainder Theoryby vyom bharadwaj - Sunday, 30 September 2007, 11:51 AM

really a grt..stuff...

was actually look ing for this one...in a compiled format...

and...AAAAHHH!!! here it is ..

thx fundoo..

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Re: Modulo Arithmetic Remainder Theoryby mon haldi - Sunday, 30 September 2007, 06:52 PM

hi AD P

u ve attempt all most correctfor 40! mod 83 until 41!^2 mod 83 =141! mod 83 = 1 or -1 thn 41!^2 mod 83 = 1 there must be some theorem 4 iti didnt got the ans bt it ll be 2 or 81TG ll he lp to find the correct solution

WHAT DO U SAY TG ABT THS PRBLM

post d solution if anyone has it

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Re: Modulo Arithmetic Remainder Theoryby parag kumar - Monday, 1 October 2007, 02:06 AM

why can't 41!%83 be -1 ?

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Re: Modulo Arithmetic Remainder Theoryby AD P - Monday, 1 October 2007, 07:21 AM

@ mon haldiWilson's Theorem

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Re: Modulo Arithmetic Remainder Theoryby mon haldi - Monday, 1 October 2007, 11:53 AM

i dnt mean to say that

41! mod 83 cant be -1

yes it can be -1 or 1

n thus i predicted ans is e ither 2 or 81

bt which one is correct i dnt knw

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Re: Modulo Arithmetic Remainder Theoryby Totalgadha fan - Monday, 1 October 2007, 01:12 PM

2 2 2 2 2 2 2 2 2 2 gud

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Re: Modulo Arithmetic Remainder Theoryby parag kumar - Tuesday, 2 October 2007, 12:01 AM

I have a big doubt on your approach. 2^4 mod 5=1 and 2^198 mod 199=1 doesn't mean that 2^396 mod 1990=1. It can even bechecked by the fact that 2^396 and 1990 are both even and hence the remainder should be even. My solution is like this -

2^1990 mod 1990 = 2^1989/(199*5)

using chinese remainder theorum,

2^1989 mod 199(A) = 114(r1)

2^1989 mod 5(B)= 2(r2)

for 199x+5y=1, x=-1 and y=40

hence remainder = 199*2*-1+114*5*40=22402

or, 22402-995*22=512

since the numerator and denominator were both divided by 2 hence the actual remainder is 512*2=1024.

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Re: Modulo Arithmetic Remainder Theoryby Total Gadha - Wednesday, 3 October 2007, 04:23 AM

Remainder when 40! is divided by 83:

We know that 82! + 1 is divisible by 83.

82! = 1 2 3 .... 39 40 41 42 ... 81 82 --> remainder by 83 = 1 2 3 .... 39 40 - 42 - 41 - 40 - 39 ... - 3 -2 - 1

= (40!)2 42 41 ---> remainder with 83 = R2 62.

Nor R2 62 + 1 is divisible by 83 --> R2 62 + 1 = 83k --> R2 = (83k - 1)/62 ---> k = 3 and R2 = 4 --> R = 2 = 2 or 81.

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Re: Modulo Arithmetic Remainder Theoryby Rishi Kapoor - Wednesday, 3 October 2007, 04:36 AM

Hey TG please look at the blog entries and reply my query that I have posted there...

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Re: Modulo Arithmetic Remainder Theoryby mon haldi - Wednesday, 3 October 2007, 06:40 AM

hi TG

bt question remains the same

what is the exact ans b/w 2 & 81

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Re: Modulo Arithmetic Remainder Theoryby Neophyte 2CAT - Wednesday, 3 October 2007, 03:25 PM

nice examples to cover a ll the concepts thank you very much

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Re: Modulo Arithmetic Remainder Theoryby Nitin Singhal - Wednesday, 3 October 2007, 04:59 PM

Hi TG,

yesteday i was struggling with a problem which ccomes under remainder theory.Please expla in as i tried to solve it with above giventheoroms but resulted in a very complex solution.Just te ll me if this problem could be solved in any simpler way.The problem is

3^1001 is devided by 1001 then what is the remainder.

Many thanks,

Nitin

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Re: Modulo Arithmetic Remainder Theoryby Varun Choudhary - Thursday, 4 October 2007, 01:23 PM



@Shubhagata roy

rem of 7^115 divided by 114??

7^3=343

7^115=7^(3*38+1)

=>7^115=(343^38)*7

So, the qstn is now (343^38)*7/114.

Also,114*3=342.i.e 1 less than 343

Therefore by remainder theorem it becomes = (1^38)*7/114 =7/114

=>Ans is 7

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Re: Modulo Arithmetic Remainder Theoryby fundoo bond - Thursday, 4 October 2007, 03:05 PM

hi nitin,1001 is a prime no. if i m nt m istak ing...so its euler no. is 1000.hence rem ll be 3?whts d given ans?

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Re: Modulo Arithmetic Remainder Theoryby Nitin Singhal - Thursday, 4 October 2007, 06:53 PM

No fundoo bond

1001 is not a prime number.It is devisible by 7,11 and 13.

The answer is 947.

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Re: Modulo Arithmetic Remainder Theoryby Total Gadha - Thursday, 4 October 2007, 10:31 PM

Hi Nitin,

1001 = 7 11 13

Now, 31001 when divided by 7, 11, and 13 gives remainders 5, 3 and 9, respective ly. Let N = 31001.N = 7a + 5, N = 11b + 3, N = 13c + 9. Multiply first equation by 22, second equation by 21.22N = 154N + 11021N = 231N + 63, Subtracting we get, N = 47 - 77k ---> 78N = 3666 - 1001MAlso, N = 13c + 9 ---> 77N = 1001c + 693

Subtracting N = 2973 - 1001L = 2002 - 1001L + 971

Therefore, the remainder is 971.

Total Gadha

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Re: Modulo Arithmetic Remainder Theoryby Ankush Jain - Friday, 5 October 2007, 02:12 AM

1001 = 77 * 13.

If we use chinese remainder theorem also,

a = 77, b = 13

r1 = 47, r2 = 9

x = -1, y = 6

Therefore remainder = axr2 + byr1 = -77*9 + 13*6*47 = 2973

Remainer (2973/1001) = 971.

So, i think 971 is the answer

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Re: Modulo Arithmetic Remainder Theoryby Varun Choudhary - Friday, 5 October 2007, 01:31 PM

@ Deep Agrawal - for 7^115 / 114.



Phi(114)=36 and not 66.... .i.e. 114(1-1/2)*(1-1/3)*(1-1/19) = 36

therefore we are left with 7^7/114 and 7^3/114=1

=>7^7=7^6.7 =>qstn becomes 7/114= ans=7

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Re: Modulo Arithmetic Remainder Theoryby Deep Agrawal - Saturday, 6 October 2007, 01:24 AM

@Varun

Thanks for pointing it out. Am sorry for the mistake

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Re: Modulo Arithmetic Remainder Theoryby Amalesh Bandopadhyay - Saturday, 6 October 2007, 01:58 PM

U r really a champ.. Thanks for the wonderful article...

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Re: Modulo Arithmetic Remainder Theoryby Ameya Raich - Saturday, 6 October 2007, 10:27 PM

Hello sir.. Could you pls te ll me how to use above identities in solving sums like : remainder of:1) 77777777.... (37 digits) mod 19 ? 2) 123412341234 ... (89 times) mod 19 ? is there anything e lse we need to know to solve these ? My friend says for Q1. we can use 7777..18 times mod 19 =1 since Euler of fi(19) =18... but how can we use this Fermat's little thm on"no. of DIGITS" ??

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Re: Modulo Arithmetic Remainder Theoryby Sageer Abdulla - Saturday, 6 October 2007, 10:31 PM

Reminder(2^1990)divided by 1990

In this case N=1990 =2.5.199so find the reminder when divided by 199 and multiply it wid 2*5 since 2 and 199 are co primes. again divide by 199 if its divisible.. thatsthe reminder..

suppose if we get 25 as reminder when divided by 199.

multiply 25 wid 2*5 ie 10.

we will get 250.. which should be divided by 199 agion to get the reminder..

this should be the simplified version of answer.

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Re: Modulo Arithmetic Remainder Theoryby rishu batra - Monday, 8 October 2007, 08:10 AM

thanx fundooo for this m indblowing tutoria l...it is of immense help...but i have a big doubt...does modular arithmetic follow these rules?:

1.(remainder of 1 expression)*(remainder of second expression)=(remainder of overall expression)

2..(remainder of 1 expression)+(remainder of second expression)=(remainder of overall expression)

3..(remainder of 1 expression)-(remainder of second expression)=(remainder of overall expression)

does this mean if 1998*1999*2000 is divided by 47,we find remainders individually with respective terms in the expression n multiplythem to get the final answer??

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Re: Modulo Arithmetic Remainder Theoryby Shashank Balooni - Monday, 8 October 2007, 11:19 AM

Hi FUNDOO .....U have really xpla ined the doubts lucidly .....Thanks a lot

I have 2 questions ...1.39! divided by 41 in this question i didnt get 40!=41K+40 ..Some more question on wilson theorem...

2.Please giv some example where we cant apply any other theorem but only Chinese Remainder Theorem



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Re: Modulo Arithmetic Remainder Theoryby Varun Choudhary - Monday, 8 October 2007, 03:01 PM

@ rishu batra

Though I was not asked but still i fe lt like answering it...

Answer to ur qstn is a simple yes..

the rule says that a ll the operators (+,-,*) remain as they are and need to be operated on the individual remainders which are obtainedby separate ly finding out remainder for each of the numerators..

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Re: Modulo Arithmetic Remainder Theoryby Shashank Balooni - Monday, 8 October 2007, 03:12 PM

Hi guys This seems quiet Diff prob ..please help

Remainder for ..by Chinese theorem(777)^777/1000

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Re: Modulo Arithmetic Remainder Theoryby fundoo bond - Monday, 8 October 2007, 03:48 PM

hi rishu,

u hv asked really nic qns n i think i shld hv included these things in the article...but anyways it's better late than never...

ur 1st statement "1.(remainder of 1 expression)*(remainder of second expression)=(remainder of overall expression)"

yes this is true...i ll take d example u hv given....1998*1999*2000 is divided by 47

find rem for 1998,it is 47.so for 1999 n 2000 it ll be 48 and 49 resp.

so rem ll be 24*25*26 % 47..

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Re: Modulo Arithmetic Remainder Theoryby sachin singh - Monday, 8 October 2007, 07:43 PM

Is the answer for this is 72???i have used euler theorem here....confirm me the answer.

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Re: Modulo Arithmetic Remainder Theoryby love kumar - Monday, 8 October 2007, 10:56 PM

Hi

there is a error in d solution of ex. to find last 3 digit of 57^802 while calculating (1000)=1000(1-1/2)(1-1/4),it shud be 1000(1-1/2)(1-1/3) as 1000=(2^3)(5^3)..

bye n tk cr

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Re: Modulo Arithmetic Remainder Theoryby Amit Chandaliya - Tuesday, 9 October 2007, 08:14 AM

hi a ll,

I find one problem while solving one of the papers.



Q. what will be the remainder if (2222^5555+5555^2222) is divided by 7 ?

plase give answer with complete solution...

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Re: Modulo Arithmetic Remainder Theoryby sukshima naik - Tuesday, 9 October 2007, 11:56 AM

thanx a lot sir.....i have not joined any coaching centre here....as i find more materia ls here from u than any coaching classes......thesefundoo theorams r not even in good books..... sir i will be appearing cat-2008 & i have started my prep. on QA now....but i need somebasics or correct planning & method to be strong in QA.SO sir can u help me to choose a correct path....

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Re: Modulo Arithmetic Remainder Theoryby Varun Choudhary - Tuesday, 9 October 2007, 01:17 PM

@ Amit Chandaliya

Hi Amit, perhaps this has been discusse in the forum before also...

2222 = 7*317+3 and 5555 = 7*793+4

=> Qstn becomes (3^5555+4^2222)%7

3^3 % 7 = 27%7 = -1

4^3 % 7 = 64)%7 = 1

=>3^5555 = 3^(3*1851+2)

and 4^2222= 4^(3*740+2)

now, we have first part of qstn as :-

((27)^1851 * (3^2)) %7 = (-1)^1851 * 2 =-1*2 = -2

and second part is :-

((64)^740 * (4^2)) %7 = (-1)^740 * 2 =1*2 = 2

Therefore answer wud be (-2)+(2) = 0

Hope this is a full solution.....

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Re: Modulo Arithmetic Remainder Theoryby Varun Choudhary - Tuesday, 9 October 2007, 01:24 PM

@love kumar

U found a mistake but u didn't look into it carefully.Its a typo(printing mistake) not an error.

The typo is dat (1000)=1000(1-1/2)(1-1/4) shud be 1000(1-1/2)(1-1/5) as as 1000=(2^3)(5^3).. => u have to take 2 and 5 as "a"and "b" respective ly.

The resultant value comes to 400 only...

so solution is alright...hope you got ur mistake....

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Re: Modulo Arithmetic Remainder Theoryby Shashank Balooni - Tuesday, 9 October 2007, 02:47 PM

Hi PLease Somebody solve this one This seems quiet Diff prob ..please help

Find the Remainder for this expression..(777)^777/1000

(Chinese theorem may help here)Show parent | Reply



Re: Modulo Arithmetic Remainder Theoryby rishu batra - Tuesday, 9 October 2007, 07:42 PM

hi a ll,

how do we solve the followin problem:

find the remainder when 123412341234.....(89times)is divided by 19?

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Re: Modulo Arithmetic Remainder Theoryby Ameya Raich - Tuesday, 9 October 2007, 09:47 PM

tens digit of 2^999 = ?

Now obviously we have to divide by 100 and find the remainder.If we use Euler, then some things going wrong and i cant fine out wat ?

See.. Rem (2^999/ 100) = 4* (Rem of (2^997)/25) ........(1)

Now :: Rem of (2^997)/25since 2 and 25 are coprime.Euler totient of 25 = 20,Rem of 2^20/25=1

Thus Rem of 2^980/25=1.Thus we have to find Rem(2^17/25)

now, 2^11/25 yield 24 as remainder, we know, thus 24* Rem(2^6/25)= Rem (24*14/25)= 11.

Now from step (1) above final remainder = 11*4 = 44.

Where am i going wrong ???????

:::::::::::::::::::::::::ALTERNATE ::::::::::::::::::::::::::

Rem (2^999/ 100) = 4* (Rem of (2^997)/25) ........(1)now, we know , rem(2^11/25)= -1thus, Rem( (2^990)/ 25) = Rem (2^11/25 * 90) =1.Thus we are left with Rem (2^7/25) = 7;now frm step (1) above , final answer is 4*7 = 28 .

Where am i going wrong ???????

?????????????????????????????????????????????????????????????

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Re: Modulo Arithmetic Remainder Theoryby love kumar - Tuesday, 9 October 2007, 09:55 PM

hi varun!!

pretty rite..

thanks

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Re: Modulo Arithmetic Remainder Theoryby RAMA KRISHNA TANKU - Wednesday, 10 October 2007, 08:04 AM

thanku u fundo , it is very useful for me , iam very much in confusion before this .Any way once again thank u

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Re: Modulo Arithmetic Remainder Theoryby monu varshney - Wednesday, 10 October 2007, 12:27 PM

As the theories given by fundoo and tG r fabulous. But I have some queries.

IN wilsons theory,how the 2nd example gives 1 remainder.pleae somebody make it clear. Treat this question as u have not done 1st example of same theorem.Also explain the Wilsons theorem morefor my convenience.

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Re: Modulo Arithmetic Remainder Theoryby Varun Choudhary - Wednesday, 10 October 2007, 02:04 PM

@Ameya Raich

you are getting wrong in the very preliminary step.

which is 2^11 % 25= -1 => wrong buddy

infact it shud be 2^10 % 25= -1

Follow the very same steps that u have taken.....

You will get the very same answer in both the case and for ur ease i wud say the answer is 88

Also u said that

Rem( (2^990)/ 25) = Rem (2^11/25 * 90) =1.Thus we are left with Rem (2^7/25) = 7; =>wrong,it shud be 3now frm step (1) above , final answer is 4*7 = 28 .

I am rewriting the above steps with corrections

Rem( (2^990)/ 25) = Rem (2^(10*99+7)/25) => (-1)^99 * rem (2^7 /25) => -3 =>22

final answer is 4*22 = 88

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Re: Modulo Arithmetic Remainder Theoryby Varun Choudhary - Wednesday, 10 October 2007, 02:49 PM

hi everybody.....

plz solve (19^36 + 17^36) %111...

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Re: Modulo Arithmetic Remainder Theoryby Naresh K - Wednesday, 10 October 2007, 05:08 PM

Thanx Bond..That was a good compilation

I fe lt Chinese Remainder Theorem to be a little redundant. Frankly we can always logically evaluate the answer in such given conditions.

In ur example, 3^101 % 77 u have arrived at the number of the form 7a+5 or 11b+3. So the remainder will be the least number commonto such series. Quick ly it will give 47 !!

Looking fwd for more to comeCheersNeo

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Re: Modulo Arithmetic Remainder Theoryby fundoo bond - Wednesday, 10 October 2007, 06:17 PM

Hi Shashank,Are u sure the qn. is (777)^777/1000 ?? I am finding it diff. to crack it...m ight be it is (777)^777/100 ?? wht do u say? Any takers for this?

but frank ly i think u wont find such pbms in CAT...

regards,fundoo

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Re: Modulo Arithmetic Remainder Theoryby Pavan Padekal - Friday, 12 October 2007, 08:37 AM



Shouldnt phi(63) in the first example be 36 and not 18 as mentioned? It doesnt make a difference to the final answer though

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Re: Modulo Arithmetic Remainder Theoryby Shashank Balooni - Friday, 12 October 2007, 11:01 AM

hi

Fundoo Bond (777)^777/1001

oops take the question as abv 1000 replaced by 1001(7*11*13)

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Re: Modulo Arithmetic Remainder Theoryby Total Gadha - Monday, 15 October 2007, 01:55 AM

1. 777777/1000 --> Remainder = 797 (Use Binomial to find the last three digits)

2. 644

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Re: Modulo Arithmetic Remainder Theoryby Anish Rai - Monday, 15 October 2007, 05:48 PM

Nice one TG

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Re: Modulo Arithmetic Remainder Theoryby HARI POTTER - Tuesday, 16 October 2007, 08:57 AM

hi haldi

can you pls expla in the remainder theorem u used...

substitution of 10^ 2 by 2

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Re: Modulo Arithmetic Remainder Theoryby HARI POTTER - Tuesday, 16 October 2007, 10:02 AM

hi TG

how do u find last n digits using binomial theorem.... atmost how many 'last' digits can we find....pls expla in how u got 797 in777^777/1000.. i learnt how to find the last 2 from your posts... but last 3 i dunno..

thanks

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Re: Modulo Arithmetic Remainder Theoryby fundoo bond - Thursday, 18 October 2007, 07:30 PM

hi shashank,

if this is the qn...

(777)^777/1001

1001=(7*11*13)

so basically u have to find remainder of

(777)^776/(11*13)

do this by chinese thm and then multiply the resultant rem by 7.

if u still need my assistance done be afraid to revert.

regards,

fundoo



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Re: Modulo Arithmetic Remainder Theoryby ashish bhardwaj - Saturday, 20 October 2007, 11:50 PM

hi varun ,

it can be solved using chinese remainder theorem

.. (19^36 + 17^36) %111 so i will divide this into two part

fist 19^36%111 and other 17^36%111

so our 111 ==> 3*37 both are prime

so a=3 now we have to find the remainder for

19^36/3 =>1=r1

19^36/37=> 1=r2 so and

our a and b are 3 and 37 so 3*x+37*y=1 it gives me x=-12 and y=1

so 3*-12*1+37*1*1= -36+37 =>1

now same way it can be done for 17^36

so 17^36/3 => 2^36/3 => 8^12/3 => (-1)^12/3 =>1

17^36/37 =>1

so it will a lso follow the same pattern and answer will be 2 =>1+1...

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Re: Modulo Arithmetic Remainder Theoryby mojo bond - Sunday, 21 October 2007, 12:00 AM

Hi Fundoo

Please clarify one thing in the question which shashank asked i.e. 777^777/1001 cancelling 7 leaves 111 extra a lso i.e it becomes(777^776 * 111) / 11*13.So altho 777^776 /11*13 is managable but how to take into account 111 ?

Plz help

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Re: Modulo Arithmetic Remainder Theoryby ambar patil - Friday, 21 March 2008, 05:49 PM

find the last three digits of 31994 .

Now as per Euler theorem , Rem [3400/1000] = 1 .

Sim ilarly , Rem [32000/1000] = 1 . Can anybody te ll me how do we proceed further ? How do we handle the extra 36 ?

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Re: Modulo Arithmetic Remainder Theoryby TG Team - Saturday, 22 March 2008, 01:40 PM

find the last three digits of 31994 .

Hi Ambar

Here I go.

32000 = 36*31994 = 1(mod1000)

Let 31994 = x(mod1000) and as 36 = 729=> 729x = 1(mod1000)=> 729x - 1000y = 1Now using Eulid's Algorithm1 = 3(1) - 1(2)1 = 3(1) - 1(8 - 2*3)1 = 3(3) - 1(8)1 = 3(19 - 2*8) - 1(8)1 = 3(19) - 7(8)1 = 3(19) - 7(84 - 4*19)1 = 31(19) - 7(84)1 = 31(187 - 2*84) - 7(84)1 = 31(187) - 69(84)1 = 31(187) - 69(271 - 187)1 = 100(187) - 69(271)1 = 100(729 - 2*271) - 69(271)



1 = 100(729) - 269(271)1 = 100(729) - 269(1000 - 729)1 = 369(729) - 269(1000)

=> 31994 = 369mod1000

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Re: Modulo Arithmetic Remainder Theoryby Vipul Patk i - Monday, 24 March 2008, 04:52 AM

Too good...Some of the problems are solved in Arun Sharma's book and it is a lways good to see alternative solutions to such k ind ofproblems.

Regards,

Vipul

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Re: Modulo Arithmetic Remainder Theoryby ambar patil - Monday, 24 March 2008, 04:50 PM

hey kamal ,

Thanks for posting the solution ... but i am not aware of the Eulid's theorem ... can you please post that too ??

However , i have found one other way to solve the problem

31994 = 9997=( 10 - 1) 997 .

now using binomial expansion , we can get the laste three digits . the answer comes outtabe 369 .

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Re: Modulo Arithmetic Remainder Theoryby Hungry Gadha! - Friday, 18 April 2008, 07:50 PM

Just Amazing JOB.....fundoo.. fantastic...stuff.

Thanks,

HG

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Re: Modulo Arithmetic Remainder Theoryby Ams Kul - Sunday, 4 May 2008, 12:18 PM

Hi Fundoo

Thanks a lot for the ar ticle ..

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Re: Modulo Arithmetic Remainder Theoryby vamshi machanapally - Sunday, 11 May 2008, 12:30 PM

Hi... I want the solution for the prblem given belowwhat is the remainder when 7^7^7 is divided by 13...

Thanks in advance

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Re: Modulo Arithmetic Remainder Theoryby vamshi machanapally - Sunday, 11 May 2008, 07:44 PM

Hi guys.. got the procedure

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Re: Modulo Arithmetic Remainder Theoryby a a - Monday, 12 May 2008, 10:01 PM

Hi vamshi

is the answer 6???

please confirm

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Re: Modulo Arithmetic Remainder Theoryby Nikunj Gupta - Monday, 19 May 2008, 09:11 AM

Hi....One thing that I am not getting in Chinese theorm is how to set values for x,y because there might be the case when there are morethan 2 values satisfying eq ax+by=1...for eg what is the remainder when we divide 123456789 by 6.....

i am not getting the ans in any case...

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Re: Modulo Arithmetic Remainder Theoryby TG Team - Thursday, 29 May 2008, 05:17 PM

Hi NikunjHere goes the explanation.Don't hesitate to ask anything which you don't follow.Q. what is the remainder when we divide 123456789 by 66 = 3*2=> 3x - 2y = 1by mere observation we get to know that x = 1 and y = 1.Now 123456789 = 1mod2also, 123456789 = 0mod3.Now using CRT,

123456789 = 3*1*1 - 2*1*0 = 3mod6.

Alternative Method:

Rem[123456789]/6 = Rem[3*41152263]/3*2 = 3*Rem[41152263]/2 = 3*1 = 3

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Re: Modulo Arithmetic Remainder Theoryby Biswajit Sen - Friday, 30 May 2008, 03:53 PM

how do you solve 38! / 41 ?

if we use the same method used to solve 39!/41, then it looks something like this.

40! / 41 = 40.39.38! / 41

Remainder [ 40! / 41 ] = Remainder [ 40.39.38! / 41]

ie. 40 = 40 . Remainder of (39*38! / 41)

that means remainder of (39.38! / 41) should be 1.

that means remainder of (38! / 41) will be 1/39 ??????

i am not convincedd.

please clarify??????

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Re: Modulo Arithmetic Remainder Theoryby akshay mishra - Sunday, 15 June 2008, 10:11 PM

hi TG!!!

plz help me wid dis problem - rem(8^182)/65

i ws doing it wid chinese remainder theorem-

65=13*5, >>>>>a=5,b=13;

rem(8^182)/5=rem((8^45))^4)*(8^2)/5)=4>>>>r1

rem(8^182)/13=12>>>r2

now 5x+13y=1;>>>x=-5,y=2;

so ar2x+br1y=212???????

i cud get da answer as follows-

phi(65)=48

rem((8^182)/65)=rem((8^48)^3)*8^38)/65

>>=rem(8^38)/65

>>rem((8^4)^9*8^2)/65

rem(8^4)/65=1

thrfre>>=rem(8^2)/65=64

but wat abt chinese remainder theorem wats wrong wid dat??



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Re: Modulo Arithmetic Remainder Theoryby Ramkrishna Roy - Wednesday, 18 June 2008, 02:07 PM

toooooooooooo good...thank you sir.

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Re: Modulo Arithmetic Remainder Theoryby praveen kumar - Friday, 20 June 2008, 04:16 PM

hi ppl,

i m anew user to this forum. N i am lik ing it. All sums r interesting. And i hv a doubt. In the sum remainder of 40!/83 i m not able tounderstand how the following step arrived from

(81.80.79.--------.42.41!)%83=1-2*-3*-4*-5....-41*41! = 41!*41!%83 = (41!)^2%83 = 1which is possible only when 41! mod 83=1because there are even no of terms so negative sign got cancelled.

Could someone expla in me this one.

Praveen.

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Re: Modulo Arithmetic Remainder Theoryby arpit m ishra - Saturday, 21 June 2008, 01:39 PM

hi neo .........this is the easy ques..........no need to use chinese theorem

8^182 = (8^2)^91 =64^91 =(65-1)^ 91 which when divide by 65 gives remainder of -1 or 64.......

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Re: Modulo Arithmetic Remainder Theoryby Biswajit Sen - Monday, 23 June 2008, 10:59 PM

hi guys......this is an interesting question for a ll to work out...very simple one.....just a small trick involved....

if A=33*32*31*30*29*28

&

if Remainder [ A/(12k)] = 0......then find the minimum value of k?

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Re: Modulo Arithmetic Remainder Theoryby Neo Sinha - Monday, 23 June 2008, 10:57 PM

K=0

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Re: Modulo Arithmetic Remainder Theoryby shivam mehra - Tuesday, 24 June 2008, 08:59 PM

plz help this part gone over head totallyLet 31994 = x(mod1000) and as 36 = 729=> 729x = 1(mod1000)=> 729x - 1000y = 1Now using Eulid's Algorithm1 = 3(1) - 1(2)1 = 3(1) - 1(8 - 2*3)1 = 3(3) - 1(8)1 = 3(19 - 2*8) - 1(8)1 = 3(19) - 7(8)1 = 3(19) - 7(84 - 4*19)1 = 31(19) - 7(84)1 = 31(187 - 2*84) - 7(84)1 = 31(187) - 69(84)1 = 31(187) - 69(271 - 187)1 = 100(187) - 69(271)1 = 100(729 - 2*271) - 69(271)1 = 100(729) - 269(271)



1 = 100(729) - 269(1000 - 729)1 = 369(729) - 269(1000)

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Re: Modulo Arithmetic Remainder Theoryby manshul goel - Thursday, 26 June 2008, 09:13 AM

really awesome fabulous and excellent ...

Thanxx Fundoo Bond..

u did a gr8 job..

Mind blowing

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Re: Modulo Arithmetic Remainder Theoryby abhinav tripathi - Thursday, 26 June 2008, 11:43 PM

thanks a lot for such a useful article...........one thing which i did'nt understand was y we have to xpress32^32 in the form 6k+r????? plz e laborate.....!!waiting for reply

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Re: Modulo Arithmetic Remainder Theoryby Surendran Chandravathanan - Tuesday, 1 July 2008, 08:13 PM

Hi,

The value of k = 1 for sure.

Bcoz 12 = 3 x 4 & we already have one 3 and one 4 in the numerator(ie, 33 = 3x11, 28=4x7).

So when A is divided by 12, sure ly the remainder will be zero.

Hence, K can be 1 since minimum value is asked.

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Re: Modulo Arithmetic Remainder Theoryby Surendran Chandravathanan - Tuesday, 1 July 2008, 09:02 PM

Hi,

The value of k = 1 for sure.

Bcoz 12 = 3 x 4 & we already have one 3 and one 4 in the numerator(ie, 33 = 3x11, 28=4x7).

So when A is divided by 12, sure ly the remainder will be zero.

Hence, K can be 1 since minimum value is asked.

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Re: Modulo Arithmetic Remainder Theoryby shubham singh - Tuesday, 1 July 2008, 10:42 PM

the minimum value of k will be 1.

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Re: Modulo Arithmetic Remainder Theoryby shubham singh - Tuesday, 1 July 2008, 10:56 PM

hi neo.u can find out the answer by the remainder theorm.the answer will be 64

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Re: Modulo Arithmetic Remainder Theoryby Gaurav Verma - Wednesday, 2 July 2008, 02:30 PM

good ,nice theorems ..i ts time saving a lot ..

please help me 2 solve this :



find the 3 digit prime no. thats divides 2000!divided by 1000!^2

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Re: Modulo Arithmetic Remainder Theoryby Rakesh Agarwal - Saturday, 12 July 2008, 11:28 AM

hi i am a new user out hereand found it a very useful forum..now Gauravi think your question is greatest 3 digit prime no. which is a factor of (2000!)/(1000!^2)..if it is so the solution is here...if you se lect a prime no. say 673now 2000! has 2000mod673 i.e 2 factors as 673 and 1000! has 1000mod673 i.e 1 factor of 673 so 1000!^2 has 2 factors of 673 so2000!/(1000!^2) has no factors of 673.so all prime no.s above 673 will fo llow the same trend..now take the case of a prime no. less than 673..671 is not a prime no. as it is divisible by 11also 667 is not e ither as it is divisible by 23so next prime no is 6612000! has 2000!mod661 i.e 3 factors of 661 which are 661,1322 and 1983since 1000! has 1000!mod661 i.e 1 factor of 661 so (1000!^2) has 2 factors of 661so we arrrive at the ans 2000!/(1000!^2) has 661 as one of its factor and is the largest prime no. that can divide 2000!/(1000!^2)hope u got that....te ll me if i m wrong.. plzregards Rakesh

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Re: Modulo Arithmetic Remainder Theoryby Pooja Sinha - Tuesday, 15 July 2008, 02:07 PM

Hey,

The last example (while explaining the concepts i.e. Find the remainder when 8^643 is divided by 132 )that wassolved had made use of Euler's theorum, whereas I tried to make use of Chinese remainder Theorum by the followingmethod :

Break 132 into 4 and 33 and since 4 and 33 are co-primes therefore find

Rem(8^643/4) = 0=r1

Rem(8^643/33) -> Applying euler's theorum we get remainder as 17=r2

Hence, Rem(8^643/132) = 4r2x + 33r1y where 4x+33y=1 (x,y) = (-8,1)

Hence, Rem(8^643/132) = 4*17*-8 + 33*0*1 = -544 , which is not the right solution.

Kindly help and point out my mistake in the approach adopted.

Regards,

Pooja

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Re: Modulo Arithmetic Remainder Theoryby amit prakash - Wednesday, 16 July 2008, 01:04 AM

hi sir...........im amit.....the way you are solving d problems is excellant...its rock ing sir....

Sir i didnt understand the Euler's theorem properly.....mainly the second line of the first example........plz expla in again........

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Re: Modulo Arithmetic Remainder Theoryby inderpreet makkar - Thursday, 17 July 2008, 06:29 AM



Hey FB

U truly are a bond w i th hell lot of fundas...

Thnx.

IPS.

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Re: Modulo Arithmetic Remainder Theoryby premdeep singh - Thursday, 24 July 2008, 09:16 AM

sir,plz give idea how to solve arun sharma solution what will be remainder 2^2+22^2+222^2 + 2222^2+................+(222........48times)^2%9

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Re: Modulo Arithmetic Remainder Theoryby premdeep singh - Wednesday, 30 July 2008, 11:22 AM

hi guy plz accept challenge cat2008 nov16 what is remainder 2^2+22^2+222^2+2222^2.......................+(222......49 times )^2%9

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Re: Modulo Arithmetic Remainder Theoryby sahil arora - Wednesday, 30 July 2008, 07:54 PM

Great work Boss..U r truly TG(total genius.... )

Fantastic,Mindblowing!!!!!

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Re: Modulo Arithmetic Remainder Theoryby sahil arora - Wednesday, 30 July 2008, 08:25 PM

hi...

could you please e laborate more in solving the above question..

(ii) remainder of 2 ^1990 / 19901990=2.5.199from Euler's Theoremphi(2)=1phi(5)=4phi(199)=198

////////// according to euler's theorm remainder of [2^phi(2) / 2] = 2^1 /2 comes out to be 1 but its zero..could uplease correct me if m interpreting it wrong and please throw some light on the part right below..that LCM and modthing ...unable to get it... ////

2^4=1 mod 5--------(1)2^198=1 mod 199---------(2)LCM(4,198)=396so2^396=1 mod 19902^1980=1 mod 1990(because 1980 is a multiple of396)so now we remain with2^10 mod 19901024 mod 1990=1024

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Re: Modulo Arithmetic Remainder Theoryby sanjeev ganesh - Saturday, 9 August 2008, 02:47 PM

Hi AD P,

Anwer to the (2^1990 % 1990) is impressive, but can you explain step LCM(4,198). Why did you take LCM ?

Sanjeev

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Re: Modulo Arithmetic Remainder Theoryby Anamica Sinha - Sunday, 10 August 2008, 02:20 AM

thanks a lot.....!!!!!!!

It's really a great help ... for we don't find these sort of stuffs in books easily.... though a bit confusing at first .....

With so many theorems to solve sim ilar Quants question .... it's a bit tough to judge which one to apply... without much practice ....

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Re: Modulo Arithmetic Remainder Theoryby sandeep shekhar - Saturday, 16 August 2008, 04:24 PM

Thanks a lot!I a lways wondered how to answer such questions.This article was wonderful.do keep posting such articles.

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Re: Modulo Arithmetic Remainder Theoryby anju mor - Sunday, 17 August 2008, 02:46 PM

PLZ te ll me some1how to solve.........1. (100/62)^1/42. (100/62)^1/53. (100/62)^1/6

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Re: Modulo Arithmetic Remainder Theoryby anju mor - Sunday, 17 August 2008, 02:47 PM

PLZ te ll me some1how to solve.........1. (100/62)^1/42. (100/62)^1/53. (100/62)^1/6

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Re: Modulo Arithmetic Remainder Theoryby abhay deol - Monday, 18 August 2008, 12:21 AM

hello TGsir do u have hard copy of your book on numbers....or do u have only the ebook option...nd sir where are the ans of these quizzes andlastly sir do your coaching classes provide any study materia l by correspondance...if someone doesnt want to attend classes....sir do reply

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Re: Modulo Arithmetic Remainder Theoryby sanjeev ganesh - Monday, 25 August 2008, 05:12 PM

I have a query.

In case of Chinnese method:

1. Do we need to check whether the number and divisor to be co-prime2. If yes, can you expla in me your last example 8^643 / 132 using Chinnese method

Please expla in

Cheers,Sanjeev



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Re: Modulo Arithmetic Remainder Theoryby sanjeev ganesh - Tuesday, 26 August 2008, 04:11 PM

Thank you for this beautiful artic le,

What is the remainder incase of 13^40 mod 49 ?

Cheers,

Sanjeev

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Re: Modulo Arithmetic Remainder Theoryby arun ilangovan - Wednesday, 27 August 2008, 05:19 PM

hi, this is my first post to TG...

(2^2 + 22^2 + 222^2...(2222...48times)^2) % 9

[(4*(1^2) + 4(11^2) + 4(111^2)...] %9

4[1^2 + 11^ 2 + 111^ 2...] % 9

4[1 + (9+2)^2 + (108+3)^2...] % 9...(a ll terms of form (a+b)^2, where a is a multiple of 9, like 9, 108, 1107...)

4[1^2 + 2^2 + 3^2+...+48^2} % 9 .........(since all terms of form a^2+2ab are divisible by 9)

[4*48*49*97/6] % 9

remainder = 5...

please let me know if this is correct...

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Re: Modulo Arithmetic Remainder Theoryby arun ilangovan - Thursday, 28 August 2008, 11:57 AM

(13^40) % 49

this can be written as...

=[(2*7-1) ^40] % 49

By binomial expansion...

=(2*7)^40-40C1 * (2*7)^39 (1)...40C39 * (2*7)(-1)^39 + (-1)^40

all the operands in the above equation except for the last two are divisible by 49. So we need to find the remainder, when the last twoterms are divided by 49.

=> -560 % 49 + 1

=7 * (40 * -2)%7 + 1

=7 * (-10) + 1

= -70 + 1

= -69

= 49*2 - 69

=29

Hence the remainder is 29....

Somebody please let me know if this is the right answer...

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Re: Modulo Arithmetic Remainder Theoryby Mukesh Kumar - Saturday, 6 September 2008, 06:14 PM

don't have words to expla in the beauty of this article..solved some difficult question with in second...thanks for this

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Re: Modulo Arithmetic Remainder Theoryby A K - Saturday, 13 September 2008, 09:49 PM



The article has made life a bit easier when it comes to remainders.

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Re: Modulo Arithmetic Remainder Theoryby nim ish vikas - Sunday, 14 September 2008, 12:16 AM

thank you fandu bondfor this compilation and you effortsnim ish

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Re: Modulo Arithmetic Remainder Theoryby anju mor - Friday, 19 September 2008, 12:19 PM

gud morning...I have some problems plz solve.

1. Find remainder when[(1^1+2^2+3^3+.....+9^9)+(10^9-11^8+12^7-13^6+14^5.....+18^1)]divided by 19.

2. First 99 natural numbers are written side by side to form a number. 1234567891011............9899find remainder when this number is divided by 11.

3. Consider a 21 digit no. 12345678........Find remainder when divided by 11.

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Re: Modulo Arithmetic Remainder Theoryby Debashish ... - Saturday, 20 September 2008, 06:15 PM

hii...anju...is d answer of d 1st question 0?? i have solved it as follows..given expression can be written as:(1^1+18^1)+(2^2-17^2)+...+(9^9+10^9)

Since,(a^n+b^n) is divisible by a+b when n is odd, (a^n-b^n) is divisible by a+b when n is even...Hence...the terms in each of the brackets above are divisible by 19...Hence the remainder is 0.

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Re: Modulo Arithmetic Remainder Theoryby Debashish ... - Saturday, 20 September 2008, 07:09 PM

3. I have solved this problem using remainder theorem and got the answer 7. is it correct?

1234...21digits=12345678910...15=(15*10^21+14*10^19+13*10^17...10*10^11)+(9*10^8+8*10^7+...1) (i) (ii)applying remainder theorem,since,11=10+1,substitute -1 in place of 10 in the above polynomial in 10...we get,as remainder,

=-(15+14+13+....10)+(9-8+7-6+...+1) (i) (ii)=-70.Now the final answer will be the result of (-70 mod 11)=7,in dis step i have written -70 as (10*70)mod 11 =7.(as remainder of -1 means

10 in this case).Dunno if i am correct[ ]...Do correct me if i'm worng sumwhere...thnkss.

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Re: Modulo Arithmetic Remainder Theoryby aarti kaushik - Saturday, 20 September 2008, 11:49 PM

Thnxs TG.

Plz solve my problem.

Q.1 What is the remainder when (1!)^3 +(2!)^3 +(3!)^3 +(4!)^3 + -----------------------(1152!)^3 is divided by 1152??????

Q.2 What is the remainder when 2(8!) -21(6!) divides

14(7!) + 14(13!)?????????????????????

Waiting for ur reply.

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Re: Modulo Arithmetic Remainder Theoryby yog bansal - Sunday, 21 September 2008, 12:58 AM

Ans 2:- 2(8!)-21(6!) = 2(8!)-3.7.(6!) or

= 2.8.(7!)-3.(7!) or

= 13(7!)

Now, remainder of 14(7!) + 14(13!) = remainder of 14(7!) + remainder of 14(13!) or

=14/13 + 0 or

=14/13 (ans.)

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Re: Modulo Arithmetic Remainder Theoryby yog bansal - Sunday, 21 September 2008, 01:15 AM

Q.1 What is the remainder when (1!)^3 +(2!)^3 +(3!)^3 +(4!)^3 + -----------------------(1152!)^3 is divided by 1152??????

Ans 1 :- 1152 = 2^7 . 3^2

Therefore, we can easily calculate the remainder of each term individually here.

e.g. a ll terms after (3!)^3 are complete ly divisible by 1152

hence, the remainder is 225 (which is equal to the remainder of the sum of first 3 terms).........Ans

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Re: Modulo Arithmetic Remainder Theoryby anju mor - Sunday, 21 September 2008, 04:00 PM

debashish thanks for d solution.........

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Re: Modulo Arithmetic Remainder Theoryby aarti kaushik - Sunday, 21 September 2008, 05:10 PM

Thnxs,

I ve another problem.

Q .1 Find the 28383rd term of the series: 12345678910111112...........??????

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Re: Modulo Arithmetic Remainder Theoryby mohit saxena - Saturday, 18 October 2008, 01:32 AM

777^777 /10001000 & 777 are coprime we apply euler th.1000(1-1/2)(1-1/5)=400

777^400 /1000 gives rem 1now left with 777^347 /1000 1000= 8*125 we divide seperate ly by 8 & 125 777^777 /8 1^777 /8=1 rem 777^777 /125

777^100 will give rem1 (aplying euler)777^700 will a lso give rem 1now left wid

777^77 /125



27^77/125

3^231/125 3^100 will give rem 1 (euler th.)

3^31 /125 (3^5)^6 *3 /125 (-7)^6 * 3 24 *3 / 125=72so rem= 72 now using chinese

8x+1= 125 y + 72x=(125y+71)/8=(5y+7)/8=y=5

put y in (125 y +72)

125*5+72=697

697 is the ans

bhai marna mat galt ho to..........

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Re: Modulo Arithmetic Remainder Theoryby pravesh bhutoria - Tuesday, 21 October 2008, 03:08 AM

hi tg,

QUES.find the last digit of the product of a ll 2 digit numbers that give a remainder of 2 when divided by 5?

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Re: Modulo Arithmetic Remainder Theoryby Software Engineer - Tuesday, 21 October 2008, 12:22 PM

pravesh, Is it 6?- SE

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Re: Modulo Arithmetic Remainder Theoryby TG Team - Wednesday, 22 October 2008, 12:07 PM

Hi Pravesh


Smallest and largest two digit number which give a remainder of two when divided by 5 are 12 and 97 respective ly.So total such numbers are = (97-12)/5 + 1 = 18 each giving a remainder two.

Hence product of last digits of a ll these 18 numbers = 29*79 = 2*7 = 4

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Hi Pravesh


Smallest and largest two digit number which give a remainder of two when divided by 5 are 12 and 97 respective ly.So total such numbers are = (97-12)/5 + 1 = 18 each giving a remainder two.

Hence product of last digits of a ll these 18 numbers = 29*79 = 2*7 = 4

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Hi PraveshQUES.find the last digit of the product of a ll 2 digit numbers that give a remainder of 2 when divided by 5?

29*79 = 2*7 = 4



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Re: Modulo Arithmetic Remainder Theoryby Software Engineer - Wednesday, 22 October 2008, 06:39 PM

of the product of a ll 2 digit numbers.......... I m issed '2 digit' part

and got 2^10 * 7^10 = 4*9 = 6

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Re: Modulo Arithmetic Remainder Theoryby pravesh bhutoria - Thursday, 23 October 2008, 01:53 AM

hi kamal,thanks for attempting this ques, but i was not able to understand the logic behind (97-12)/5 + 1 such numbers and how come

there product will be 29*79. so will u pls expla in it thoroughly?

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Re: Modulo Arithmetic Remainder Theoryby pravesh bhutoria - Thursday, 23 October 2008, 03:45 AM

hi tg sir,

i have a conceptual doubt.what if, in chinese reainder theorem,

N=a*b*c. That is N has 3 prime factors.can the chinese remainder theorem still be applied.if so then how?what about ax+by=1 in thatcase?

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Re: Modulo Arithmetic Remainder Theoryby TG Team - Thursday, 23 October 2008, 03:51 PM

Hi PraveshLet's take an example:

What'll be the remainder when 1111...2008 times is divided by 1001?

Now let the number is N.Also 1001 = 7*11*13

So N = 4mod7 = 0mod11 = 11mod13 .Using. 2*(11) - 3*(7) = 1=> N = (2*11*4 - 3*7*0)mod77 = 11mod77

Using, (13)*6 - (11*7) = 1=> N = (13*6*11 - 11*7*11)mod1001 = 11mod1001.

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Re: Modulo Arithmetic Remainder Theoryby Pushpender Pannu - Tuesday, 31 March 2009, 10:12 PM

hello can anybody help me of 63 is given as 18. It should be 36?

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Re: Modulo Arithmetic Remainder Theoryby Ankita Chowdhury - Wednesday, 1 April 2009, 08:33 AM

Hi Pushpender,

63=3^2*7

phi(63) = 63(1-1/7)(1-1/3) = 36*6/7*2/3 = 36

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Re: Modulo Arithmetic Remainder Theoryby sanjeeb panda - Saturday, 4 April 2009, 08:51 AM

Thanks Sir for these tricks and examples..

P ls sir write as much as possible becoz it he lps to us those who are unable to take or admit into your class room coaching.

Thanks Sir.



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remainder of 2 ^1990 / 1990?by kundan kumar - Tuesday, 21 April 2009, 02:12 AM

please expla in were m i goin wrong

2^1989/9952^1980 * 2^9/9952^1980 * 2^9/5*199[now by eulers theorem 2^1980 leaves remainder 1 when divided by 199 ( phi(199)=198) ]therefore we r le ft with 2^9/5which leaves remainder 2

but reading the above thread 1024 is the correct ans...

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Re: Modulo Arithmetic Remainder Theoryby Vibhor Goyal - Sunday, 10 May 2009, 01:40 PM

Hi Kamal/TG sir

I have a doubt in the solution provided for the remainder when 1111...2008 times is divided by 1001. In the 5th line of the solution, the

remainder when 1111...2008 times is divided by 7 is taken as 4. The value as per my calculation is 5..can you pls te ll me where im making a

mistake??what i did was:with 1 -> 111 -> 4111 - 61111 - 511111 - 2111111 - 01111111 - 1and so on..hence for 2008 times digit 1, remainder should be 5.rgds

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Re: Modulo Arithmetic Remainder Theoryby Gowtham Muthukkumaran Thirunavukkarasu - Wednesday, 13 May 2009, 05:47 PM

how to solve (i) find the remainder of 55555 .... 93 times divided by 98

55555.........93 times = 5( 1+10+100+........+10^98)= 5((10^99)-1)/ (10-1)therefore,the reminder when the number is divided by 98 is,= 5*10^99/(9*98) - 5/(9*98)= 0

I hope that this is right. If the number was prime then u could use the concept that any single digit number repeated 'n' times is divisibleby 'n' provided 'n'is prime.

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Re: Modulo Arithmetic Remainder Theoryby Ashutosh Singh - Monday, 18 May 2009, 12:49 PM

Really a very helpful and effective article . An eye opener

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Re: Modulo Arithmetic Remainder Theoryby vakati babu - Thursday, 28 May 2009, 09:35 PM

for the very first question

5^37 / 63

how f(63)=63*(1-1/3)*1-1/7=18 . it is 36

how it is confirmed that

5^18/63 remainder is 1

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Re: Modulo Arithmetic Remainder Theoryby vakati babu - Wednesday, 3 June 2009, 12:09 AM



in euler's theorem can some one please expla in the calculation of the fucntion pi( n)

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Re: Modulo Arithmetic Remainder Theoryby Nabanshu Bhattacharjee - Wednesday, 10 June 2009, 11:59 PM

I approach such problems, in which the numerator and the dinominator has some common factors, in the following way

1) take the common factors out i.e. divide both numerator and dinominator by their HCF

2)Find the remainder( now the numerator and the dinominator would be co-primes)

3) multiply the remainder with the HCF

So in this problem, we can write

21990 = 2* 21989

1990=2*5*199

phi(5) =4

phi(199) =198

LCM(4,198) =396

therefore 2396 = 1 mod 995

therefore remainder when 2^1989 divided by 995 is 29

hence the required remainder is 2*29

I think this clears a lot of cloud on the theory being applied here

I am new here...correct me if I am wrong

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Re: Modulo Arithmetic Remainder Theoryby Nabanshu Bhattacharjee - Thursday, 11 June 2009, 12:21 AM

1001 is a beautiful number. any three digit number multiplied by 1001 results in the number written twice i.e. abc * 1001 =abcabc

in the prob 1111...2008 times is divided by 1001

111111 is divisible by 1001( 111 * 1001)

that means 111.....2004 times would be divisible by 1001

(111111* 1000001000001000....etc)

that would leave us with 1111

the remainder would be 110

@pravesh

I think the best approach in such cases is by tak ing LCMs, and then getting a smaller number to divide

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Re: Modulo Arithmetic Remainder Theoryby Nabanshu Bhattacharjee - Thursday, 11 June 2009, 12:24 AM

Have a crack at this

remainder when 72! is divided by 73*36!

(I dont know where this problem appears. I got this from a friend through SMS. Hopefully I am not infringing any copyright )

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Re: Modulo Arithmetic Remainder Theoryby Sai Shyam Kolachalama - Saturday, 13 June 2009, 09:48 PM

Thanks a m illion ... I a lways wanted to know how to solve these k ind of problems... ... TG .. u Rock ... .... and fundoo... u

rule ..



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Re: Modulo Arithmetic Remainder Theoryby neha lahoti - Sunday, 14 June 2009, 01:53 AM

Hi TG

i used some other method to solve 32^32^32 when divided by 9, by my method i am getting remainder =2 when 32^32^32 is divided by9.

we can write the question as 2^5^32^32 where 5^32^32 can be wriiten as 3k +1 where k is an even number.

so now 2^3k+1 divided by 9 will give you 2 as the remainder.Could you please point out my mistake.

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Re: Modulo Arithmetic Remainder Theoryby Kulvir Singh - Sunday, 14 June 2009, 12:35 PM

@ NehaJust think about this question and you will get where you are wrong..(8)^2=(2^3)^2=2^(3*2)=2^6Now what you did wrongly was writing the given question as 2^5^32^32 where as it should have been (2^5)^32^32 = 2^[5*(32^32)]

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Re: Modulo Arithmetic Remainder Theoryby neha lahoti - Monday, 15 June 2009, 12:29 PM

Hi Kulvir.

Thank you for the solution.I have a doubt which is:

(111..7) times whendivided by 7 will give a remainder 1 and if (111..14 )times will a lso give the remainder as 1 when divided by 7.?Is thiscorrect.

Could you expla in me the logic behind the chinese theorem. I mean how is it derived since thr has to be a logic behind...thanks

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Re: Modulo Arithmetic Remainder Theoryby Kulvir Singh - Tuesday, 16 June 2009, 01:40 PM

@ nehaIts an extension of the actual theorem stating that 111111 will be complete ly divisible by 7...So,use it in this form only..This automaticallyanswer your second question that 111...(14 times) will actually give a remainder of 11 i.e. 4 with 7..It's better form of the same theoremna..??

I dont know the exact proof,but i can te ll you the logic behind this..10^6 will give me a remainder of 1 with 7(fermet's theorem)Hence 10^6-1 should be divisible by 7..Hence 999999 should be divisible by 7..9*111111 must be divisible by 7...Since 9 is not divisible by 7,111111 should be divisible by 7..And hence 222222,333333 and so on will a ll be divisible by 7..The same can be extended by to a ll prime numbers..

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Re: Modulo Arithmetic Remainder Theoryby neha lahoti - Tuesday, 16 June 2009, 05:56 PM

Hi TG

Thanks...This site is 2 gud 2 b true!!1 more doubt!

#how many pairs of x and y ex ist for which the

eq: sqrt X + sqrt Y= Sqrt 1332 holds true.Could you plz solve it.It is one of the questions from TG quizzes.Are solutions given to the quizzproblems on the site? Thanks

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Re: Modulo Arithmetic Remainder Theoryby Kulvir Singh - Wednesday, 17 June 2009, 09:07 AM

@ nehaThe answer will be 5...The solution are not there in that quiz...Actually i would suggest TG to make the solutions available because thatway we can learn various aproaches to solve the problems on various concepts...[And hey!!You can even charge some chhoti c amount for

that... ]



Well now tak ing over the businessman part of me,the solution for this question...rhs of the question after factorisation gives me sqrt(36*37)That can be written as 6sqrt(37)Now ek chhota ca concept...If the rhs of the equation has sqrt of 37,both sqrt(x) and sqrt(y) shud have sqrt(37) in it...Now the question reduces to a sqrt 37 + b sqrt 37= 6 Sqrt 37Hence a+b=6(a,b) can be 1,5 ;2,4; 3,3; 4,2; 5,1And subsequently,value of X and Y can be found out...Hope i am clear...!!

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Re: Modulo Arithmetic Remainder Theoryby Kulvir Singh - Wednesday, 17 June 2009, 09:42 AM

@ remainder when 72! is divided by 73*36!The answer will be 27*36!

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Re: Modulo Arithmetic Remainder Theoryby Nabanshu Bhattacharjee - Saturday, 20 June 2009, 04:09 PM

How did you get it?

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Re: Modulo Arithmetic Remainder Theoryby Antonio Banderas - Wednesday, 24 June 2009, 03:37 PM

good article..cheers

can anyone help me with this??

find 28383rd term of the following 123456789101112...

options are 3,4,9,7..

my answer is coming 3..but its wrong

did it by tak ing 1-9-->9*1digits

10-99-->90*2=180digits

100-999-->900*3=2700digits

total=2889digits of 1,2 and 3 digit numbers.

so, 4digit numbers will have 28383-2889=25494digits

so the actual number will be 999+[25494/4]=7372

while dividing with 4 we get remainder 2 so the required digit is going to be 3 as we'll have 7 3 7 2 7 3 7 3...and so on

what am i doing wrong??

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Re: Modulo Arithmetic Remainder Theoryby kartik jani - Thursday, 25 June 2009, 02:37 PM

by the way... 63(1-1/7)(1-1/3) = 36 not 18 as u mentioned after introducing the first theorem

k indly make the necessary changes if possible... a very nice article thoughmore of these expected....

a lso wat will be the remainder if 34! is divided by 73...

some quick and helpful method pls

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Re: Modulo Arithmetic Remainder Theoryby Ahefaz Khan - Thursday, 25 June 2009, 09:16 PM

This is by far the best article I have ever and will ever come across. I am a business journalist and just wonder why are you people tak inga measured approach and not coming out in a big way when every tom, dick and harry is resorting to every possible gimmick to reach outto CAT aspirants.

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Re: Modulo Arithmetic Remainder Theoryby Kulvir Singh - Friday, 26 June 2009, 09:18 AM

34! is divided by 73The answer will be 36...No quick method to get this one...You will have to invest a good 3,4 minutes to reach to the answer..

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Re: Modulo Arithmetic Remainder Theoryby Kulvir Singh - Friday, 26 June 2009, 09:23 AM

@ Antonio BanderasI guess you have seen this question in Arun sharma were in the answer m ight be given as something other than 3...But your approach andanswer is correct...Even i am getting the answer to be 3 only..I think the answer is wrong..Kindly te ll from where you have taken thisquestion so that the credibility of answer be verified..

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Re: Modulo Arithmetic Remainder Theoryby Antonio Banderas - Friday, 26 June 2009, 01:22 PM

yeah its from Sharma..where the answer is 9..

here 's another sum

p=1!+2*2!+3*3!+........+12*12!

what'll be the remainder when p+3 is divided by 13!?

my approach is this

for any number we can use the following

when we have 1!+2*2!+3*3!+.....+(n-1)(n-1)! this form and it is divided by n! the remainder is -1...

(eg of -ve remainder: if we have 23/6, we can express 23 as 6*4-1(6k-1)

or we can express 23 as 6*3+5(6k1+5) )

eg: as in 1!+2*2!+3*3! when divided by 4! will give 23/24...

here remainder is 23 which we can write as 24k-1...so i write the remainder as -1...thus 23=4!-1

thus in the given sum p/13! will give remainder as -1 and 3/13! will give remainder 3..

thus the remainder should be 3-1=2....

i hope my method is ok..i don't have the answer to this.

if anyone gets a different answer to this or has a different method please do share...

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Re: Modulo Arithmetic Remainder Theoryby srinivasan ravi - Saturday, 27 June 2009, 12:31 PM

hi a ll,anyone pls help me to solve this problem,

if x = 777...777 ( 101 7 are there )then what is x mod 440 ?thanks..

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Re: Modulo Arithmetic Remainder Theoryby Antonio Banderas - Saturday, 27 June 2009, 01:09 PM

well here x will be divisible by 440 if it is divisible by 4,11,10(440=4*11*10).

11's divisibility rule is (sum of odd digits)-(sum of even digits)=0/11k

here if we take 98 7's then we have the following

777777......98digits000+777

the number 77777....98digits000 is divisible by 11(49*7-49*7=0)

is divisible by 4(ending with 3 0's)

is divisible by 10.

so answer should be remainder of 777/440 =(777-440)= 337

is this the answer??

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Re: Modulo Arithmetic Remainder Theoryby Kulvir Singh - Saturday, 27 June 2009, 02:12 PM

Your approach to both the questions is correct...And both the answers are also correct..

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Re: Modulo Arithmetic Remainder Theoryby srinivasan ravi - Sunday, 28 June 2009, 08:36 PM



thank you antonio..this is the answer..thanks a lot..

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Re: Modulo Arithmetic Remainder Theoryby Nabanshu Bhattacharjee - Monday, 29 June 2009, 01:55 AM

Kulvir,

about the prob 72! divided by 36! 73

The approach i followed is,

take 36! out from both the numerator and denominator,

The new numerator, 37.38.39.....72 ,let us call it A,can be written as (73-36)(73-35)(73-34) etc,

So A% 73 is same as 36! % 73., let us call it x

Hence the Answer to the problem is 36!x

Now, 72!/73 is -1 or 72

i.e. 36!.A %72 = -1 or 72

i.e x^2%73 is -1 or 72

1.e x^2 = 73k - 1

I too got 27 by inspection, but that is too tedious. P lease share your approach.

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Re: Modulo Arithmetic Remainder Theoryby abin abraham - Monday, 29 June 2009, 09:17 PM

Just Marvellous most remainder questions had me stumped even before i cd start think in on them nw i ve got rt d concepts

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Re: Modulo Arithmetic Remainder Theoryby Kulvir Singh - Monday, 29 June 2009, 11:43 PM

Well i a lso did something sim ilar but a bit smaller. But before going into the solution,i would like to put forth that such questions are veryless like ly to come in an exam like CAT because there is no short method to solve it. Its just a matter of an epiphany.Well here is my approach..72!mod73 = -1(72*71*70*.......36*35*34*.....*2*1)mod73 = -1(-1*-2*-3*-4........-35*-36)*(36*35*....2*1)mod73=-1(36!*36!)mod73=-1(36!)^2mod73=-1Now here is the catch..To me,it struck by chance that since 27*27=729 gives a remainder of -1 with 73, (36!)mod73 will be 27..Hence 34!*35*36mod73 = 2734!*19 mod 73=2719a=73b-27From here,value of b will be 9...And hence,a is 36..Quite a long solution actually...

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Re: Modulo Arithmetic Remainder Theoryby arvind kumar - Thursday, 2 July 2009, 12:13 PM

beautifull expla ination thank yuou so s o much.

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Re: Modulo Arithmetic Remainder Theoryby Ching Lee - Friday, 3 July 2009, 12:45 PM

@ Gowtham Muthukkumaran Thirunavukkarasu

with regard to ur solution dated 13 May,2009....



Could u pls explain the last step of ur answer i.e.

Remainder of 5*10^99/(9*98) - 5/(9*98)= 0How come?

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Re: Modulo Arithmetic Remainder Theoryby Ching Lee - Friday, 3 July 2009, 12:52 PM

@ Nabanshu Bhattacharjee

could you pls throw some more light on the LCM approach that you have mentioned in one of ur posts. There is only1 example which everyone has repetitively explained for the LCM approach. Could u give me any other examplewherein u hv used this approach.

@ all

or maybe someone else has other examples.

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Re: Modulo Arithmetic Remainder Theoryby Sidhant Bansal - Friday, 17 July 2009, 04:24 AM

thanx a lot TG.... awesome stuff !!

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Re: Modulo Arithmetic Remainder Theoryby Manish Kumar - Saturday, 18 July 2009, 05:50 PM

Hi,Gr8 article...

Had few doubts in the above explanation of example in Euler's Theo:

1) How come 63(1-1/3)(1-1/7)=18 and not 36?

2)1000(1-1/2)(1-1/4)=400 and not 375?

Thanks,Manish.

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Re: Modulo Arithmetic Remainder Theoryby tarun bhavnani - Friday, 24 July 2009, 02:03 PM

just did the exercise.. u ve made finding remainder so easy...thnx TG...

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Re: Modulo Arithmetic Remainder Theoryby Astha Jalan - Monday, 27 July 2009, 10:53 AM

Hey,Ho do we find the remainder of 12^107 / 37 ?thanks!

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Re: Modulo Arithmetic Remainder Theoryby ronak pate l - Monday, 27 July 2009, 05:59 PM

hi.. is that ans 1? by euler's theorem phi(37)=36

therefore rem[12^36/37]=1now rem[12^107/37]=rem[{12^(36)*3}/37]*rem[12^-1/37] =1*rem[1/12*37] =1 am i right?if wrong somewhere make me notice it...

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Re: Modulo Arithmetic Remainder Theoryby shalin ja in - Monday, 27 July 2009, 08:23 PM

--find the remainder of 55555 .... 93 times divided by 98

This question was asked by Mr. Subhadeep das, 2 yrs back, on 28th September 2007. I think it had been lost in the "deluge" of TG

followers' queries .

First of a ll,Mr. Gowtham, this remainder cannot be 0 as the Nr. is ODD and Dr. is even.

We can do it this way..111111 is divisible by 49.So 111111.... 90times will a lso be divisible.So,[5(111111...90times*1000) + 555] / 98

1st part of the addition will be divisible by 98 complete ly as 1111...90times part is divisible by 49 and 1000 is divisible by 2.

Now when 2nd part 555 is divided by 98, 65 will remain..

I think 65 shoud be the answer.

Correct me if i m wrong.

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Re: Modulo Arithmetic Remainder Theoryby aashish bia la - Tuesday, 28 July 2009, 02:54 PM

hi TG ,

its amazing work you are doing. in this world of money makers you are doing socia l service and that too when you know this is the onlyway your work is going to be applauded. Good job done TG.

keep up the good show and cheers!!!!!!!

thanks

Aashish Bia la

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Re: Modulo Arithmetic Remainder Theoryby swati ja in - Monday, 10 August 2009, 04:56 PM

Hi TG ..

Wannna know how do we evaluate remainder whn question is like

a/b^n .. Can it be solved by chinese remainder theorm ?

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Re: Modulo Arithmetic Remainder Theoryby aditya kaul - Tuesday, 1 September 2009, 02:36 PM

I guess you are going in the last step. If you take the negative powers then any power of 12 can henceforth be proved to have aremainder 1.I am not so good with theorems and hence cant give you a correct reason but what I can give you is a corect solution

rem[12^36/37]=1

now in this way rem[12^108/37] = 1

this can be written as..rem[12^107/37 * 12/37]= 1

now rem[12/37] can be written as 12 or -25

now lets say rem[12^107/37] is xthen rem[x * -25/37] should be equal to 1this will happen only if x = -3 as rem[75/37] =1

hence rem[12^107/37] = -3 or 34

Ans-34

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Re: Modulo Arithmetic Remainder Theoryby multi loop - Wednesday, 2 September 2009, 04:04 PM

Hi Mate. Kudos to you for putting up this article. It presents the requisite stuff in a



comprehensible manner.

I wonder why you haven`t included the 'Method of Splitting the Divisor' in your text..

For instance consider this one (This is a question from the test series of one of the leading

instiutes).....

Qn. N = 7777.....7 upto 2n+1 digits, n>3. Find the remainder when it`s divided by 1232.....?

Sol. 1232 = 7 x 11 x 16

N = 7p = 11q + 7 = 16r + 1 for some values of p,q,r.

Trying to solve 7p = 11q + 7... As 11q should be a multiple of 7, we try q = 0 and get p = 1.

So, Rem [N/77] = 7 i.e. 77p + 7 = 16q + 1 or 16q - 77p = 6

p = 2 and q = 10 satisfy it. Hence the least such number is 16(10) + 1 = 161.

So, N = 77(16)k + 161.

I failed to understand this solution i.e. why would it give the remainder.

Please make me understand it.....

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Re: Modulo Arithmetic Remainder Theoryby Ashwin Agra

Chinese Remainder Theorem

Documents

Transcript of Chinese Remainder Theorem