Concepts for CATH

8/11/2019 Concepts for CATH

1/56

A Collection by ymafreak@PG Page 1

Contents

Concepts Total Fundas posts of maxximus .................................................................................... 2

#31............................................................................................................................................. 2

#35............................................................................................................................................... 4

#18............................................................................................................................................... 5

#41............................................................................................................................................... 8

#322.......................................................................................................................................... 10

#52............................................................................................................................................. 14

#59............................................................................................................................................. 16

#60............................................................................................................................................. 17

#323........................................................................................................................................... 19

#325........................................................................................................................................... 21

#230.......................................................................................................................................... 23

#76........................................................................................................................................... 26

#103........................................................................................................................................... 33

#120........................................................................................................................................... 35

#120 Solution ........................................................................................................................ 37

#129........................................................................................................................................... 38

#127........................................................................................................................................... 39

#132........................................................................................................................................... 41

#137........................................................................................................................................... 44

#135 Solution #132 ................................................................................................................... 45

#344........................................................................................................................................... 47

#345........................................................................................................................................... 48

#192........................................................................................................................................... 49

#193........................................................................................................................................... 50

#Solution 193 ........................................................................................................................ 53

#408........................................................................................................................................... 56


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Concepts Total Fundas posts of maxximus

http://www.pagalguy.com/forum/quantitative-questions-and-answers/23536-concepts-total-fundas.html

post# 1,18,31,32,35,41,52,59,60,76,80,103,112,120,127,12 9,132,145,163,

192,193,226,246,250,314,322,323,325,344,345,346,34 9,355,372,390,404,410,411,412,414.

#31

Here is something i used and doubt if many are aware of... This is a method to divide quickly (very Quickly )and get results with amazing accuracy...

But let me forewarn you, it needs good amount of practice before you can even think of trying it in exam hall...

And yes, this is not my creation. Credits to Rahul and company who shared this with me 2 yrs ago

so here we Go!!!

The approach would be to get the denominator to either a 100 or a 1000because that is what percentages is

all bout.

Simply focus on the fact that how do the given denominators reach 100/1000. I have left some blank becausethey are very obvious.

The following are the stations between 100 and 1000

100

111 - Reduce 10%

125 - Multiply by 8133 - Reduce 1/4

150 - Reduce 1/3

166 - Multiply by 6182 - Add 10%

200

222 - Reduce 10%

250273 - Add 10%

300

333 - Reduce 10%364

400

455 - Add 10%

500555 - Reduce 10%

600
http://www.pagalguy.com/forum/quantitative-questions-and-answers/23536-concepts-total-fundas.htmlhttp://www.pagalguy.com/forum/quantitative-questions-and-answers/23536-concepts-total-fundas.htmlhttp://www.pagalguy.com/forum/quantitative-questions-and-answers/23536-concepts-total-fundas.htmlhttp://www.pagalguy.com/forum/quantitative-questions-and-answers/23536-concepts-total-fundas.html


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666 - Add Half700

750 - Add 1/3

800

833 - Add 20%875 - Add 1/7

900

910 - Add 9%950 - Add 5%

1000

So Task (1) you have to mug up the values of these stations. It is very important that these values are

memorized because this will help you in

knowing which number to reach from any given number. e.g. if the denominator is 887 you know you have to

reach for 900 or 875 and so on.

Task (2) Practice!

Below is an approach to tackle (three digit /three digit) with consummate ease. We shall attempt to understand itwith examples.

Example 1:What is 145/182 ------------------79.5

Steps1. add 10% to numerator and denominator...

2. it becomes 159/200 ...............which is 79.5

( the answer from the calculator is 79.6)

Example 2:123/178....???

Step1:- which station is closest to it?.............................200?...or some say 150. Either is good. (identification

took 2 seconds)

Step 2:- what do I have to do to go from 178 to 200/150..................add 22/subtract 28... (another 2 seconds.)

Step 3:- so if I add 22................i am actually adding slightly less than 13% of 178 to itself.

This part is tricky.. here is how I got it:

This is how u need to think---Our number is 178. Thus, 10% of it is 17.8

22 definitely more than 10% .

if 10% = 17.8 then 5% = 8.xx

And 2.5%= 4.xxx (don't even bother to calculate xxx)

which means its around 12.5........or 13 or 12..........


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Add same to numerator...........10% of 123 = 12.3

1% = 1.23 so 2% = 2.46

So 12% = 15?So it becomes 138/200 = 69%?

(calculator answer is 69.1).

All you require to calculate is what is 10% 1%...and approx stuff...any damn calculation works in less than 8

seconds.:satisfie:

Some thing like (2456*4567 - 2134*3214)/2134*3214will taken 10 seconds maximum...

it works coz I had that time 1.5 years ago..and I aint kidding..

Lets take one more.

Example 3:532/745?

Tough???

This's how it can be approached:

# Nearest station.........750.................so add 5..............about 1% or less..add 1% damnit# Numerator now is 537.............(added 1%)

# Fraction is 537/750 add 1/3 each...........it is 71 something.

Rememberhere don't even attempt to do...537/3. Because denominator is a 1000 and not a 100. So one digit is

redundant. So all I do is........53/3 approx..18...plus..53 = 71%..(will do if answer are spaced...)

answers not spaced?Then 537 +537/3 = 537+179 = 716 which makes it 71.6 (calculator is 71.5)

Dont worry if the last few statements were difficult to digest. Try solving a few questions and you'll get the cruxof it.

Remember in CAT we don't find answers... we choose them!

Happy Computing

Libocta

#35

Quote:

Can you orally find out cube roots? I know many of you may know this, but for the sake of those who do not...lets say cube root of 328509??????

Fisrt of all lets find out whats the cube of all single digit numbers1 cube = 1


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2 cube = 83 cube = 27

4 cube = 64

5 cube = 125

6 cube = 2167 cube = 343

8 cube = 512

9 cube = 729now, from the question we see that last digit of the number 328509 is 9

so the number whose cube is 328509 should end with 9 (since only cube of 9 ends with 9)

also, 50 cube = 12500060 cube = 216000

70 cube = 343000

our number lies between 60 cube and 70 cube, so the 1st digit has to be 6, and we already have the last digit as 9

so the number is 69.

Whoever undertakes to set himself up as a judge of truth and knowledge is shipwrecked by the laughter of the

Gods

Finding out smal lest no. which l eaves specif ic remainders with specif ic divi sors.

#18

Type # 1.

find smallest no. other than k, that leaves remainder k when divided by w,x,y...

to solve such questions, take lcm of w,x,y...and add k to it.

e.g. find Smallest no. other than 4, that leaves remainder 4 when divided by 6,7,8 or 9...

take lcm of 6,7,8,9 and add 4

i.e. 504 + 4 = 508

Type # 2

find smallest no. that leaves remainder 3,5,7 when divided by 4,6,8 respectively.

unlike last case, this time the remainder is not constant. but if u see carefully, difference b/w divisor n remainder is constant. i.e. 4-

3=6-5=8-7=1

in such questions, take lcm of divisors n subtract the common difference from it

here, the answer wud be lcm of 4,6,8 i.e 24 - 1 = 23

Type # 3


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Smallest no. that leaves remainder 3,4,5 whn divided by 5,6,7 respectively and leaves remainder 1 with 11,

we have just seen a way to tackle the first 3 conditions...the no. wud be lcm of 5,6,7 - 2 = 208

now we have one more condition...remainder 1 with 11.

concept => to a no. if v add lcm of divisors...the corresponding remainders dont change.

i.e to 208, if v keep adding 210 ... the first 3 conditions will continue being fulfilled.

so, let 208 + 210k be the no. that will satisfy the 4th condition...viz (208 + 210k)% 11 = 1

208%11 = 10

210k%11 = k

therefore, 10 + k shud leave remainder 1 when divided by 11.

hence, k = 2. and the no. is 208 + 210 x 2 = 628

e.g. find the smallest no. that leaves remainder 2 when divided by 3,4 or 5 and is divisible by 7

for first 3 conditions....no. is 120 + 2 = 122

hence, 122 + 120k is the required no. which reduces to 3 + 2k when divided by 7...now 3+2k shud be a multiple of 7...easily, k=2 and

the required no. is 122 + 120 x 2 = 362

Type # 4

What if there is no relation between divisors n remainders?

e.g. find the smallest no. that leaves remainders 1 with 5, 4 with 7, 6 with 11 and 7 with 13.

we can c...there's no relation among these divisor-remainder sets...neither is the remainder constant...nor is the difference b/w divisor n

remainder a constant.

in such cases...take 1 case n target another case...e.g. i take the case 7 with 13...and target 6 with 11.

which is the smallest no. that leaves 7 with 13? 7 itself...right? so all nos of the form 7 + 13k will give 7 rem with 13.

now am targeting 6 with 11...so i divide 7 + 13k by 11...i get remainder 7 + 2k...now 7 + 2k = 6,17,28,39,50...so that the remainder

with 11 is 6.

a no. that gives integral value of k is 17 i.e. 7 + 2k = 17. hence, k =5 and the no. that satisfies these two conditions is 7 + 13 x 5 = 72

now that 2 conditions are fulfilled, lets target a third condition...say 4 with 7.

to 72, if v add lcm of 11, 13 i.e 143, 2 conditions awready satisfied wud continue being satisfied...

hence the no. is of the form 72 + 143 k.


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72 + 143k % 7 = 2 + 3k

now 2 + 3k shud be = 4,11,18,25,32... to satisfy the condition of 4 rem with 7..

a no. that gives integral soln is 11..i.e. 2 + 3k = 11, k = 3.

hence, the no. that satisfies all 3 conditions is 72 + 143 x 3 = 501.

now if v see carefully...4th condition...remainder 1 with 501 has already been satisfied...so the no. v have been looking for is 501.

For ease of calculation, start from biggest divisor n gradually move to smaller ones...u'll always see that last 1-2 conditions wil

be satisfied automatically.

there are theorems for solving above questions...viz chinese theorem etc...but i solve such questions by the way i've suggested...i find

this approach very practical as the flow of nos. is very much visible...n i believe i can tackle any twist in the question devised by cat

makers thru this method...there are lotsa other questions based on this concept which i'll soon post but the basic concept remains the

same...

few points to be noted

*you can always re-check ur answer

**at times, u can use options to solve such questions.

***dont let concepts go away believing such questions can be easily dealt with thru options...the question may not always be find the

smallest no. which...... at times it may be ..find the sum of integers of smallest no. which leaves remainders...blah blah...

****there may be a case when they put an option which satisfies all the conditions but is not the smallest poss value...n put another

option...our favorite...none of these!!! lets not undermine genius of cat makers!!

questions for practice...

find the smallest no. that leaves remainder (s)

Q1. 2 when divided by 3,5,6 or 9 (other than 2)Q2. 2,5,7 when divided by 7,10 and 12 respectivelyQ3. 1,2,3,4 with 3,4,5,7 respectively.Q4. 6 with 7,8,9,10 and 3 with 11.Q5. 3 with 6, 0 with 11, 3 with 5, 7 with 8 Q6. 2 with 5, 7 with 8, 3 with 4, 5 with 7&11 Q7. 1 with 11, 4 with 5, 9 with 10, 7 with 9.

hope the post helps...puys...come up with ur own...sweet methods...which are confined in ur sharp brains...i'll also appreciate if u cancome up with feedbacks/suggestions...like today i thought its better to give answers a day late...

regardsmaxximus


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#41Type#5

Smallest no. that must be subtracted from 1000 so that the resultant no. leaves remainders 1,3,4,8 with

divisors 2,6,5,13 respectively.

from yesterday's approach, v can find out...the smallest no. satisfying all 4 conditions is 99. now to 99 if v add

lcm of 2,6,5,13 i.e. 390, the remainders will remain unchanged.

so v need 99 + 390k such that the resultant value is just below 1000.easily, for k=2, we get one such value....99+390x2 = 879.

hence, ans is 1000-879=121

Type#6

smallest no. that leaves remainders 3,2,4 when successively divided by 5,6,7 respectively.

for such questions...start approaching from the rear end...

we want 4 remainder with 7...the smallest such no. is 4 itself.

now this 4 must have come after a no. was divided by 6

so the no. must have been 4x6+2(remainder with 6) = 26

now, 26 was the quotient when sum no. was divided by 5

so the no. must have been 26x5 + 3(remainder with 5) = 133

so, the answer is 133.

Type#7

a no. leaves remainder 3 when divided by 5 and remainder 8 when successively divided by 11. what is the

remainder when this no. is divided by 55?

look at this question carefully...55 is lcm of earlier divisors 11,5... in such a case...the remainder with lcm as

divisor wud be constant.

an easy approach for this problem...start from the rear end...take a small no. that leaves rem. 8 with 11...lets take

8.

this 8 is quotient when the main no. is divided by 5...it also leaves remainder 3.

hence, the main no. is

8x5 + 3 = 43

43%55 = 43 answer.


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Type#8

Find the largest no. that leaves same remainder when it divides 3398 and 6578.

the concept is very simple...to leave same remainder...difference between two dividents must be divisible by the

divisor.

i.e. 6578-3398 = 3180 shud be divisible by the divisor to leave same remainders.

largest no. that divides 3180 is 3180 itself.

hence, the answer is 3180.

Type#9

Find the largest no. that leaves same remainder when it divides 16009,9009,7509 and 14009.

the approach is same... take difference of the nos in ascending or descending order....i.e. 16009-14009=2000,

14009-9009=5000

9009-7509=1500.

now to leave same remainder, each of the interval shud be divisible by the divisor.

hence, take hcf of 2000,5000,1500. i.e. 500

so, the answer is 500.

Type#10

If a no. is divided by 15, it leaves a remainder 7, if thrice the no. is divided by 5, then what is the

remainder?

options...1,2,3,4,0

such questions are difficult to frame as one has to find a pattern b/w divisors n remainders...i know these

questions are easy n v all can crack it easily...but the reason y am putting it here is bcoz i have a very

short...practical approach for solving this question..

choose a no. that leaves 7 remainder with 15....lets take 7 only.

thrice 7 = 21

21%5 =1 (edited after vani's post)

since, the no. shud give same result for all values that give 7 rem. with 15, its better to take sum value n solve it


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instead of takin an algebraic approach...

hence, answer is 1.

Answers for yesterday's questions...

Q1. 2 when divided by 3,5,6 or 9 (other than 2)----92

Q2. 2,5,7 when divided by 7,10 and 12 respectively----415Q3. 1,2,3,4 with 3,4,5,7 respectively.----298Q4. 6 with 7,8,9,10 and 3 with 11.----20166Q5. 3 with 6, 0 with 11, 3 with 5, 7 with 8----1023Q6. 2 with 5, 7 with 8, 3 with 4, 5 with 7&11----2007Q7. 1 with 11, 4 with 5, 9 with 10, 7 with 9.----529

kudos to vani for the active participation n gett'n most answers correct.

Today's questions...

1. smallest no. that must be added to 1000 so that the resultant no. leaves remainders 2,3,4,5 with 5,6,7,11

respectivelt

2. smallest no. that leaves remainders 1,2,5,6, when divided successively by 2,3,4,23.

3. smallest no. that leaves remainders 4 everytime when successively divided by 7,5,10,13 respectively.

4. a no. leaves remainders 2,5,3,7 when successively divided by 3,7,6,9. what is the remainder when this no. is

divided by 126?

5.find the largest no. that leaves same remainder when it divides 2345,7645,9845,6595 and 10095.

6.a no. when divided 88 leaves remainder 3. what is the remainder when its divided by 11?

7.a no. when divided by 391 leaves rem. of 49. find the remainder when its divided by 39...options 29,10,noneof these,cannot be determined.

i'll appreciate if these questions are discussed in the thread.

with warm regardsmaxximus

#322

Remainder questions can be broadly divided into 2 types...

1). LCM based questions. e.g. a no. leaves remainders 3,2 when successively divided by 5,6...what will be theremainder when this no. is divided by 30?


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2). Power based questions. e.g. remainder when (31)^[(373)^(432)] is divided by 7.

we have already discussed type 1 in two parts...those who have missed it or wish to revisit the concept may use

the link below.

first half

http://www.pagalguy.com/forum/quanti...-fundas-2.html (Concepts...total fundas!!)

second half

http://www.pagalguy.com/forum/quanti...-fundas-5.html (Concepts...total fundas!!)

so what we're left to discuss is type 2 mentioned above...i think there are 3 ways of solving these questions...v

gotto use our own sensibility to c which approach suits where...

a) Using binomial theoremb) using cyclicity with remainders

c) using euler's theorem.

for better understanding, i'll divide this post into 3 parts...n will discuss one approach in each part...

Finding remainders using binomial theorem.

(x+y)^n can be expressed as :

nC0 x^n + nC1 x^(n-1)y^1 + nC2 x^(n-2)y^2 + .......... nCr x^(n-r)y^r + ..... nCn y^n

concept:first term, i.e nC0 x^n is the only term that is independent of y and last term, i.e nCn y^n is the only term that isindependent of x. rest, all the terms are divisible by both x & y.

we'll leverage this property to solve complex problems.

e.g. 28^37 % 9 = ?

see...v can express it as (27 + 1 )^37 % 9. now, since 27 is a multiple of 9, the only term that'll be independent

of 9 will be the last term i.e. nCn 27^0 * 1^n = 1.

hence, the remainder is 1%9 = 1.

important observation:we know 1^ (any damn thing...even infinity) = 1 and (-1)^(anything) = 1 or -1 with even and odd values of

power respectively. so we'll try getting a form of (nD +/-1)^N. so that v r ultimately left with (+/-1) N.

342^423 % 7 = ?
http://www.pagalguy.com/forum/quantitative-questions-and-answers/23536-concepts-total-fundas-2.htmlhttp://www.pagalguy.com/forum/quantitative-questions-and-answers/23536-concepts-total-fundas-2.htmlhttp://www.pagalguy.com/forum/quantitative-questions-and-answers/23536-concepts-total-fundas-5.htmlhttp://www.pagalguy.com/forum/quantitative-questions-and-answers/23536-concepts-total-fundas-5.htmlhttp://www.pagalguy.com/forum/quantitative-questions-and-answers/23536-concepts-total-fundas-5.htmlhttp://www.pagalguy.com/forum/quantitative-questions-and-answers/23536-concepts-total-fundas-2.html


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step1:

342%7 = 6.

so the question becomes...

6^423 % 7.

express 6 as 7 - 1.

so the expression becomes...

(7-1)^423.

the only terms independent of 7 is ... -1^423 = -1.

hence, the remainder is -1 + 7 = 6.

523^325 % 7 =?

=> 5^325 % 7.

now, i seek a remainder of 1 or 7-1=6 with a power of 5.

5^1 % 7 = 5....wont work

5^2 % 7 = 4....wont work.

5^3 % 7 = 6...will work.

so, (5^3)^ (325/3) = (5^3)^ (108 ) x 5

= (-1)^108 x 5

= 1x5 =5...the remainder.

529^700000 % 7

= 4^700000

4^3%7 = 1. hence, we'll express the given expression in terms of 4^3

4^3 ^ (700000/3)...c how to save time...700000%3 = (7 + 0 + 0 + 0 + 0 + 0) % 3 = 1.hence (700000-1)/3 will

be an integer...dont evaluate its value... bcoz 1^ anything = 1.

hence v have 1^I x 4^1 = 1 x 4 = 4


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therefore, the remainder is 4.

(31)^[(373)^(432)] % 7 = ?

31 % 7 = 3

hence,

=(3)^[(373)^(432)]

3^3 = 27 = 28 -1. hence, expression shud be in terms of 3^3

= 3^3 ^ [(373^432)/3]

now treat [(373^432)/3] as a new, different question...

(373^432)%3 = 1.

hence, the term becomes 3^3^(I) x 3^1

I = (373^432 -1)/3 and is of the form (odd-odd) / odd = even/odd = even for sure!!

hence, the expression becomes....

(3^3)^(even I) x 3 % 7

= (-1)^even integer x 3 = 1 x 3 = 3

hence, remainder = 3.

case when divisor is a large no.

in such questions, v try to reduce power by increasing the value of base and bringing it close to a multiple of

divisor.

e.g. 2^35 % 61 = ?

v know 2^6 = 64

hence, (2^6)^5 x 2^5

2^6 % 61 = 3.


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hence, 3^5 x 2^5

= 3^4 x 3 x 2^5

=3^4 % 61 = 20.

3x 2^5 = 96, 96%61 = 35.

hence, 20 x 35 % 61 = 700%61

= 29, the reqd answer.

hope the concept helps...will shortly post the remaining 2 parts as well...

regards

maxximus

#52Today's concept...few tricky questions on divisibility

concepts...

a^n - b^n is always divisible by a-b

a^n - b^n is divisible by a+b when n is even.

a^n + b ^n is divisible by a+b when n is odd.

a^n + b^n + c^n +... is divisible by a+b+c+.... when n is odd.

when last n digits of a no. are divided by 2^n, the remainder is same as the remainder when the entire no.

is divided by 2^n.

Answers to yesterday's questions.

1. smallest no. that must be added to 1000 so that the resultant no. leaves remainders 2,3,4,5 with 5,6,7,11

respectivelt----667


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2. smallest no. that leaves remainders 1,2,5,6, when divided successively by 2,3,4,23.----such a no. doesnt existas 5 rem with aint possible.

3. smallest no. that leaves remainders 4 everytime when successively divided by 7,5,10,13 respectively.----872

4. a no. leaves remainders 2,5,3,7 when successively divided by 3,7,6,9. what is the remainder when this no. is

divided by 126?----80

5.find the largest no. that leaves same remainder when it divides 2345,7645,9845,6595 and 10095.----50

6.a no. when divided 88 leaves remainder 3. what is the remainder when its divided by 11? ----3

7.a no. when divided by 391 leaves rem. of 49. find the remainder when its divided by 39...options 29,10,none

of these,cannot be determined.---cannot be detrmnd

kudos to rockeezee who got most of them right!

questions for today...

1. 32^23 + 17^23 is definetly divisible by....

a. 49 b. 15 c. 49 & 15 d. none of these.

2. 32^23 - 17^23 is definetly divisible by....






5. 32^2321 + 17^2321 + 13^2321 + 18^2321 is definetly divisible by....

a. 40 b. 20 c. 80 d. all of these. e. none of these.

what is the remainder when 42527152653425416242624272427215287 is divided by :


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6. 16

7. 32

8. 64

regards

Maximums

#59

Quote:

Originally Posted byjunoonmba

Hey, Can anybody tell me the logic behind the following-----

To check no. divisible by 4 or not ,check last two digits

To check no. divisible by 8 or not ,check last three digits and so on.........

The logic should be the same for the last funda given by maxximus.............

Why dont the digits from left hand side matters..??

@rockeezee,gk1,deep,vani... thankx for the participation n appreciation guys...keep it up...the former atleast...

@junoonmba....the logic is simple....if a no. is 54325387...u can express it as 54325300 + 87....

u can again express it an 100 x 54253 + 87.

now 100 x 54253 is definetly divisible by 4 as 100 is divisible by 4.

so effectively, remainder depends on 87%4.

similarly for divisibility with 8,

express the above no. as 54325x 1000 + 387.

1000 is divisible by 8...so 387%8 is effectively the remainder.

similarly...for finding remainder with 2^n, divide last n digits by 2^n.

regardsmaxximus
http://www.pagalguy.com/forum/quantitative-questions-and-answers/23536-concepts-total-fundas-6.html


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#60

Today's concept...finding factors of a no. n sorting relevant questions.

concepts...

#1. to find total no. of factors of a number.

=>first of all...express it in terms of prime numbers

e.g 1260^4.

express it as (2x2x3x3x5x7)^4 = 2^8 * 3^8* 5^4*7^4.

=>now add 1 to the powers of every prime no. n multiply them all...u get the total no. of factors of 1260^4.

i.e (8+1)*(8+1)*(4+1)*(4+1)= 2025 = total no. of factors of 1260^4.

#2 to find no. of odd factors of a number.

leave power of 2 and multiply powers of all other prime nos after adding one to them.

i.e. (8 + 1)*(4+1)*(4+1) = 225 = no. of odd factors of 1260^4

#3 to find no. of even factors of a number.

take the difference of total factors and odd factors.

i.e. 2025 - 225 = 1800. = no. of even factors of 1260^4

#4 to find in how many ways can a given no. be represented as product of two relatively prime factors.

in this case, the power of prime no. becomes irrelevant as all the powers shud lie with the same factor else thetwo factors wont be relatively prime.

i.e in the above case 1260 = (2x2x3x3x5x7)^4 = 2^8 * 3^8 * 5^4*7^4.

now powers 8,8,4,4 have no importance...whats important is how many prime nos. are there...they are four...viz

2,3,5,7. hence 4.


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now v have to form nos. using these 4 primr nos. its a like a question askin u...if u have 4 sweets...in how many

ways can u eat them? the answer is 4C0..when u eat none... + 4C1 when eat any one...+4C2 + ...4C4...when u

eat all.

similarly here, answer wud be 4C0 + 4C1 + 4C2 ...4C4 = 2^4 = 16.

but the factors have to exist in pairs...hence 2^4/2 = 8 factors are possible.

on a general note...a no. formed of n prime nos has 2^(n-1) pair of co-prime factors

Answers to yesterday's questions


a. 49b. 15 c. 49 & 15 d. none of these.


a. 49 b.15c. 49 & 15 d. none of these.




a. 49 b. 15 c. 49 & 15d. none of these.

5. 32^2321 + 17^2321 + 13^2321 + 18^2321 is definetly divisible by....

a. 40 b. 20 c. 80 d. all of these. e. none of these.

what is the remainder when 42527152653425416242624272427215287 is divided by :

6. 16----7

7. 32----23


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8. 64----55

kudos to irevani, gk who got all correct...so too to deep, rockeezee...who got jus 1 wrong....bhaiyon...if a

no. is divisible by 80...its obviously divisible by 40 n 20.

questions for today...

on the very simple fundas above, very tricky questions can be framed.

Find the smallest no. that has exactly.....

1. 16 factors

2. 12 factors

3. 60 factors

4. how many factors of 27000 are perfect cubes?

how many A.Ps are possible such that first term is 1235 and the last term is 3535 and there are atleast...

5. 3 terms

6. 4 terms

7. 5 terms

8. 6 terms

How many values of a are possible if x^2 + ax + 2400 has...

9. integral roots

10. roots which are natural nos.

regards

maxximus

#323


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Finding remainders using cyclicicty with remainders:

This approach is useful when the divisor is small or at times when it is a factor of 100.

3^327%7 = ?

3^1 % 7 = 3

3^2 % 7 = 23^3 % 7 = 6

3^4 % 7 = 4

3^5 % 7 = 4x3 % 7 = 53^6 % 7 = 5 x 3 % 7 = 1

3^7 % 7 = 1 x 3 % 7 = 3

remainder with first power is same as remainder with 7th power...hence v can say that cyclicity in remainders is7-1 = 6.

so, 327 % 6 = 3,

hence, effectively, the remainder is 3^3% 7 = 6

326^524 % 9 =?

326 % 9 = 2

hence, 2^524

now,

2^1 % 9 = 2

2^2 % 9 = 42^3 % 9 = 8

2^4 % 9 = 7

2^5 % 9 = 5

2^6 % 9 = 1.2^7 % 9 = 2

hence cyclicity in remainder = 7-1 = 6.

524 % 6 = 2

hence, the remainder is 2^2 % 9 = 4.

81^502 % 100


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81^1 % 100 = 8181^2 % 100 = 61

81^3 % 100 = 41

81^4 % 100 = 21

81^5 % 100 = 0181^6 % 100 = 81

hence, cyclicity = 6-1 = 5.

502 % 100 = 2.

so the reqd remainder is same as that with 81^2 = 61.

similarly...this method can be effectively used when remainders are other factors of 100.

viz, when the factors are 20,25,50,100...knowing last 2 digits wud suffice knowing the remainders...we'vealready discussed this concept while discussing cyclicity...

important: dont try this method when the divisors are complex...viz 37,73 etc...the cylicty wud come very late n

calculations will grow cumbersome...when divisors are complex, there must be sum other catch in thequestion...look for that catch...e.g. last type discussed in the binomial method...

regards

maxximus

#325Finding remainders using Euler's theorem.(special thanks to junoonmba for this)

This method is very useful when the divisor and dividend are relatively prime numbers...

step 1: To calculate euler's no. of a divisor.

euler's no. can be practically taken as cyclicity in remainders by a divisor..

to find euler's no, express the divisor in terms of prime factors...

100 = 2^2 x 5^2.

powers of the prime nos. have no significance...its jus the prime no. that matters...

euler's no (e for convenience) = divisor x (1-1/first prime factor) x (1-1/second prime factor) x ... (1-1/last prime factor)

so, for 100, e = 100 x (1-1/2) x (1-1/5) = 100 x 1/2 x 4/5

= 40.

that means e for 100 = 40. or, in other words, 100 divisor will definetly show a cylicity of 40 in the remainders.

whenever the power of a relatively prime no. will be a multiple of 40, the expression wud show a remainder 1 with 100.


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e.g. 3^120 % 100 = ?

we know e for 100 = 40.3 n 100 are relatively prime nos.hence, 3^40 % 100 = 1.

hence 3^120 % 100 = (3^40)^3 % 100 = 1^3 = 1.

7^100 % 45 = ?

45 = 3x3x5

e for 45 = 45 x (1-1/3) x (1-1/5) = 24

hence, 7^24 % 45 = 1

hence, 7^100 % 45 = 1^4 x 7^4 % 45

= 2401 % 45

= 16, the required answer...

with this, we have finished the conceptual part of remainders...now what we seek is practice...we need to solve a wide variety of

problems to improve our reflexes while choosing the best method for solving a question....

a sincere request to everybody around...please drop in the most difficult questions on remainders u've eva encountered...let' s

solve them thru various approaches and discuss our solutions...for the coming few days...lets know remainders inside out...

regardsmaxximus


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#230

questions to be further discussed

1. is the no. of flowers in the garden = 3576876767576654544334538?

st 1. no. of flowers in a row = no. of flowers in a column.

st. 2 no. of rows = no. of columns.

2. for above question


st 2. no. of flowers are even.

3.If p^q = 289, (p-q)^(q/2) = ?

stmt 1. p > q

stmt 2. q is not = 1. (edited...made q is not= 1, hint for all...whether its p not =1 or q not = 1, answer

remains unchanged)

4. for above questn.

stmt 1. p>q

stmt 2. p is prime.

5. is X > Y?

stmt 1. 2/5 X > 3/8 Y

stmt 2. 3/7X > 15/29 Y

i hope we get them all soon...there's sum catch in every question....so come up with ur answers as well asxplanations...

regards

maxximus

1. Going by the basics

Suppose we have r no. of rows and c no. of columns and they have f1, f2 flowers each respectively...

So total no. of flowers, in the garden = r X c X f1 X f2

Statement 1 : no. of flowers in a row = no. of flowers in a column

keeping f1=f2=f (say), we get, r X c X f^2, which may be or may not be equal to

3576876767576654544334538, hence statement 1 alone is not sufficient.

statement 2 : no. of rows = no. of columns, that gives us r = c = z(say)

hence we get r X c X z^2, which again may be or may not be equal to 3576876767576654544334538 , hence


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statement 2 alone is also not sufficient.

Now, considering both the statements together...

f1=f2=f and r=c=z, hence we are getting (fz)^2, which is a perfect square and hence can not have 8 as the last

digit ( credits to CAT 2007 ) and hence the number in question cannot be the solution...

Hence the answer should be 3

Are we missing out something maxximus?? ??:

2. again statement 1 is not sufficient alone as explained above,

statement 2 alone is also not sufficient, for obvious reasons...

keeping statement 1 and 2 together we get the no of flowers to be r X c X f1 X f2, which is an even quantity, so

both of them together also are not sufficient. Hence the answer is 5

3. statement 1 : p>q, now the p, being 17 and q being 2 and also p being 289 and q being 1 are few possibleoptions, that satisfy statement 1, but gives different (non-unique) solution for the equation in question.... hence

statement 1 alone is insufficient...

statement 2 : q is not = 1, so few potential values or p and q are

p=17, q=2p=289^4, q=0.25, etc...

which give non-unique solution to the equation in question, hence statement 2 alone is also not sufficient...

putting both the statements together, we get possible values as

p=17, q=2p=289^4, q=0.25,

which, as explained for statement 2, are not sufficient to decide. hence the answer is 5

4. statement 1 : p>q, which as explained in the previous question is not sufficient alone.

statement 2 : p is prime, now there can be many possible values of p and q, i.e

p=3, q = (log 289/ log 3)

p=17, q = 2 etc...

which do not give the same result for the equation, in question.... hence statement 2 alone is insufficient

putting both the statement together, we get, p>q and p is a prime

eg. p=5, q=3.52

p=17, q=2


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again, these values do not give the same answer for the equation, so both the statements together are also not

sufficient.... Hence the answer is 5

5. statement 1 : 2/5 X > 3/8 Y, or X > 15/16 Y, which does not tell whether X is greater than Y or not, forexample X can be 15.5 and Y can be 16 and also X can 17 and Y can be 16. So statement 1 alone is not

sufficient.

statement 2 :3/7 X > 15/29 Y, or X > 35/29 Y,

possible values to staisfy statement 2, X = 36, Y = 29, => X > Y

X = -1.19 , Y = -1, => X < Y

hence statement 2 alone is not sufficient...

Now keeping both statements together

first statement X > 15/16 Y or X > 0.9375 Y

second statement X > 35/29 Y or X > 1.207 Y (approximately)

Case 1: X and Y both are positive, the two statements imply

X > 1.207 Y , meaning X > Y

Case 2 : X is positive, and Y is negative,

X > 0.9375 Y

X > Y (A positive number is always greater than a negative number)

Case 3: X is negative, Y is positive

The two statements are not satisfy, hence its not a possibility

Case 4: X and Y both are negative, the two statements imply

X > 0.9375 Y, meaning X > Y

Hence the answer should be 3 :smile:

I know that lots of the question(s) will be wrong, please tell the correct approach(es) for them....

ThanksMohit

"Democracy, is an anarchy, having multiple dictators"

--Unknown


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Moi Blog:http://mohitranka.wordpress.com

Moi Trainee PaGal now...

#76

Today's concept...to solve tricky questions based on no. of factors of a number.

Despite the interest shown in the concept yesterday, not many cud solve the question efficiently...so, best concept for the day can be to

discuss yesterday's problems...have a quick, practical solution to them...and practice a lot of similar, different questions on same

concept. here we go...

Find the smallest no. that has exactly.....

1. 16 factors

2. 12 factors

3. 60 factors

questions are lil tricky...but if u get the concept...they become child's play...see how...

16 factors...that mean product of (powers +1) of all the prime nos = 16.

now 16 can be achieved in following ways...

16

8x24x4

2x4x2

2x2x2x2.

by sheer common sense, v can say the highest power shud go to the smallest prime no. i.e. 2....and as v proceed,

smaller powers shud be given to higher prime nos.

Concept:

powers shud reduce and the corresponding prime nos. shud increase.

so...

16...the no. is 2^15
http://mohitranka.wordpress.com/http://mohitranka.wordpress.com/http://mohitranka.wordpress.com/http://mohitranka.wordpress.com/


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8x2...the no. is 2^7 x 3

4x4...the no. is 2^3 x 3^3.

2x4x2...the no. is 2^3 x 3 x 5

2x2x2x2.....the no. is 2x3x5x7.

first three nos can be easily discarded as they are too big...just calculate last two nos, they are 120 and 210...120

is smaller and hence the answer.

important:please avoid craming...in few cases last way might give best answer..in other cases, 2nd last

one...its always advicable to form patterns n check the closer ones.

2. 12 factors

12 = 12 or 4x3 or 2x2x3. easily answer wud be 2^3 x 3^2 or 2^2 x 3 x 5

the corresponding values are 72 n 60. hence, the answer is 60.

similarly, Q3 also.

4. how many factors of 27000 are perfect cubes?

i realy wonder y nobody cud get this right...u just need to form combinations and check which

combinations give u cubes...they are 27000,27,1000, etc. am not discussing this question...i hope when i

give a similar question today...i get few correct answers.

This question is still open for answers/discussion (so are others...but if u can answer this...with an

xplanation, it'll be gr8 )


5. 3 terms

6. 4 terms

7. 5 terms

8. 6 terms

concept:


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to have first n last term as the given terms...the difference b/w the two terms shud be divisible by the commondifference. so u basically have to find how many such common differences exist...for every common

difference...u'll get a new AP.

for practice, lets take a small interval [1,15]

the difference is 15-1=14. now 14 is divisible by 1,2,7,14...four different integers... so v can have four different

APs...if v take a common difference other than these four values...the last term wont lie in that AP.

e.g if v take the common diff = 4, the AP wud be 1,5,9,13,17...see 15 didnt lie...

so, 4 APs are possible.

now if a conditions is attached...there shud be atleast threeterms...it means that AP with 2 terms shud be

neglected...1,15 is an AP with jus 2 terms...so it shud be beglected...remember, an AP with 2 terms always lies in any

interval.

so the answer wud be 4-1 = 3

if the condition is atleast fourterms...then the AP with 2 terms as well as the AP with 3 terms shud be

neglected.

we know that an AP with 2 terms is bound to exist...lets c if an AP with 3 terms also exists.

an AP with 3 terms will look like... 1, x, 15

see...there are 2intervals... x-1 and 15-x. hence for a 3 term AP to exist, the difference shud be divisible by 2.

since 14 is divisible by 2, we further reduce the answer by 1...so APs with atleast 4 terms are 3-1 =2

now, if the conditions is...atleast 5 terms...v need to check if AP with 4 terms exists...such an AP wud luk like...

1, x, y, 15

see, there are 3 intervals...since 14 is not divisible by 3, such an AP does not exist. so the answer remains 2.

similarly, for atleast 6 terms, v check if 14%4 = 0...since no, the answer is again 2

for 7 terms 14%5 is not 0, the answer is 2

for atleast 8 terms...14%6 is not 0 so the answer is again 2.

for 9 terms, 14%7 = 0. hence answer becomes 2-1 =1

keep on proceeding like this...the soln wont be so bulky...its been done like this for ease of understanding...for

ease of calculation...see how to proceed...


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5. 3 terms

6. 4 terms

7. 5 terms

8. 6 terms

3535 - 1235 = 2300.

2300 = 23x2^2x5^2 . hence, no. of factors = 3x3x2 = 18. (check yesterday's concept if missed)

how many APs...18.

how many with atleast 3 terms?

since 2300%1 = 0, 18-1 = 17


2300%2 = 0, hence 17-1 = 16


2300%3 =/ 0, answer remains 16. where =/ means not equal to


2300%4 = 0, 16-1 = 15

how many with atleast 7 terms?2300%5 = 0, 15-1 = 14


2300%6 =/ 0 , answer remains 14


2300%7 =/ 0, answer remains 14

and so on...

to check for atleast n terms, v need divisibility till n-2...i'll appreciate if u dont cram...but understand

it...i've neva learnt it...its jus an observation...

so, the answers to above 3 questions wud be 17,16,16,15.


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finally...

How many values of a are possible if x^2 + ax + 2400 has...

9. integral roots


this is an actually tricky problem...sad that nobody came up with this...see...

concept:

to solve ax2 + bx + c =0,

we break it as ax2 + mx + nx + c = 0, such that m*n = a*c.

here, a*c = 1*2400 = 2400. so v need to find in how many ways can 2400 be expressed as product of 2 nos.

every such pair of nos. will give a new value of the coefficient of x.

2400 = 2^5 x 5^2 x 3

no. of factors = 6x3x2 = 36.

but these factors have to exist in pairs...e.g when v use one factor 2 (to express 2400 as 2x1200) the other

factor...1200 is automatically used...so total pairs possible are ...

36/2 = 18.

but relax...this is not it....again...equal no. of negative pairs exist...i.e. 2x1200 corresponds to -2 x -1200.although the product is same as the reqd product i.e 2400...the sum is different...its 1202 n -1202...n v need to

find different values of sum...hence the answer wud

18x2 = 36 again...

9. integral roots


9. for integral roots...the answer wud be 36

10. values wud be 1/2 the total possible values as negative roots aint allowed...36/2 = 18 is the answer

Kudos to rockeezee...who got 5 correct answers...so too to junoonmba,vani for their active

participation!!! wud love to see all of u gett'n of the following correct today...


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Today's questions...

smallest no. that has exactly.

1. 20 factors

2. 36 gactors

3. 30 factors

how many APs are possible such that the first term is 2454 , last term is 4254 and there are atleast...

4. 5 terms

5. 7 terms

6. 10 terms.

how many of the factors of 640000 have

7. perfect square roots.

8. perfect fourth roots

9. perfect cube roots

how many different values can 'A' take if x2 + Ax + 2500 has ...

10. integral roots

11. negative roots.

12. non-negative integral roots.

13. find the no. of factors 15! has (here comes the season of fresh concepts...which'll derive from older ones...hehe)

14. find the no. of factors 18! has

Minimum value that A can take if x2 + Ax + 900 has...

15. Integral roots.

16. Negative integral roots.

17. Positive integral roots.


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this is a long post...with several concepts and a wide variety of questions...i believe it'll take some time to

grasp, solve n discuss the concept properly...so i wont post a new concept tomorrow...i've got messages

suggesting i shud be a bit slow....so lets discuss this very concept till v master it...start solving guys...at any

point of difficulty...i'll be there...

Happy solving!!!!!

with due regards

maxximus


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#103

Hie...sorry guys...was out whole day...so a late reply...

before anything else, i shud give the answers...

smallest no. that has exactly.

1. 20 factors---240

2. 36 factors----1260

3. 30 factors----720

how many APs are possible such that the first term is 2454 , last term is 4254 and there are atleast...

4. 5 terms----33

5. 7 terms----31

6. 10 terms.----29

how many of the factors of 640000 have

7. perfect square roots.----18

great work gk1, rockeezee...here's a more practical approach...640000=5^4x2^10

express this as 25^2 x 4^5. now since 25 as well as 4 are perfect squares...combination of their products willalso be pefect squares...no. of combinations here wud be (2+1) x (5+1) = 18

8. perfect fourth roots----6

express it as 16^2 x 625^1 x 4...since 16, 625 have perf fourth roots...so will their cominations...i.e. 3x2 = 6 of

them.

9. perfect cube roots----8

express it as 8^3 x 125^1 x 10...since 8,125 are perfect cubes...their combinations wud yield perf cubes...theirno. are... (3+1) x (1+1) = 8

how many different values can 'A' take if x2 + Ax + 2500 has ...

10. integral roots----16


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2500 = 2^2 x 5^4. hence no. of factors are 3x5 = 15.

no. of pairs = {15 / 2 } = 8. {} = least integer function...plz check...rounded it off to higher value...the value is

odd because of 50 x 50 = 2500...giving only one combiantion seen both wat...while other combinations give 2

different cominations when the order is changed...for example...250 x 10 and 10 x 250... but 50 + 50 will give anew sum 100...so it shud be counted.

so the answer in above case wud be {15/2}x2 = 8x 2 =16.

for better understanding...lets take one more example with smaller value of a x c.

e.g. x2 + Ax + 4 = 0. A can +/- (2+2 or 4 +1)...so 4 combinations...while 4 = 2^2...hence having 3 factors...toget the answer, we'll perform the following opeartn...{3/2} x 2(for negative counterparts) = 4.

11. negative roots----infinitely many

great job rockeezee...its not mentioned negative integral roots...so infinitely many... i know many of u tuk out

answer bcoz of faith in my typing errors...n must have thought that i forgot to type integer...well, in that

case..the answer is 16/2 = 8. as 0 cannot be a root of above expression.

12. non-negative integral roots----8

13. find the no. of factors 15! has ---- 12x7x4x3x2x2

14. find the no. of factors 18! has---- 17x9x4x3x2x2x2

...great job rock, junoon...try coming up xplanations as well...18! has 16 2s, 8 3s, 3 5s, 2 7s, 1 11, 1 13, 1 17.

added one to the no. of times each prime no. existed and multiplied them...

Minimum value that A can take if x2 + Ax + 900 has----

15. Integral roots---- (-901) while taking -900 x -1 = 900

16. Negative integral roots----60...thru 30 x 30

17. Positive integral roots--- -901thru -900 x -1

kudos to rockeezee,gk1,junoonmba,iyervani!!! who got many right.....

hope this post clears few concepts....wud suggest u all to kindly go thru this post n the original post once more

to cement the concept...

wud love to hear from u guys to make this thread more effective...as per the feedbacks so far, am trying to make

it slow n steady...

with due regards

maxximus


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#120

Today's concept : Finding HCF n LCM of typical values.

#1 : to find the HCF, LCM quickly.

i know most of us wud know this....if v have few nos., 20,40,50,80,180...to find their LCMs, HCF...there's a slightly quick method...

express them in prime nos.

20 = 2^2 x 540 = 2^3 x 550 = 5^2 x 280 = 2^4 x 5180 =3^2 x 2^2 x 5

now to HCF, see highest power of all prime nos. that are common to all nos.

2 - 23- 05 - 1

hence hcf is 2^2 x 5 = 20

to find lcm...see highest power of all prime nos across all nos.

2 - 43 - 25 - 1

hence, lcm = 2^4 x 3^2 x 5 = 1620.

#2 To find HCF and LCM of the form-

2222....30 times.

3333....70 times.

to solve such questions...

for HCF..

take hcf of no. thats being repeated...i.e. hcf of 2 & 3. i.e. 1

take hcf of no. of time these nos. are being repeated...i.e. hcf of 30 n 70...thats 10.

so the hcf is 111...written 10 times.

For LCM...


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take lcm of no. thats being repeated...i.e. lcm of 2 & 3. i.e. 6

take lcm of no. of time these nos. are being repeated...i.e. lcm of 30 n 70...thats 210.

so the hcf is 666...written 210 times.

#3...to find hcf and lcm of following form...

2^300 - 1, 8^250 - 1.

the idea is..a^n - b^n is always divisible by a-b. so v need to find highest a-b that will divide a^n - b^n and smallest term that'll be

divisible by a^n - b^n.

express them in a common base.

2^300 - 1 and 2^750 -1.

to find hcf...

take hcf of powers i.e. hcf of 300 and 750...i.e. 150

so the hcf is 2^150 - 1.

to find lcm....

take lcm of powers i.e. lcm of 300 and 750...i.e. 1500

so the hcf is 2^1500 - 1.

Questions :

find hcf and lcm of:

1.2222...250 times and 8888...300 times

2. 333....120 times and 1111...400 times

3. 111...700 times and 9999...200 times.

4. HCF of 33333...200 times. and 777777.....300 times

5. 32^250 -1 & 16 ^ 100 - 1.

6. 81^100 -1 & 243 ^ 200 - 1.

7. 343^150-1 & 2401^100 - 1.

8. 125^200 - 1 & 625^120 - 1.

9. 169^320 - 1 & 32^160 - 1.


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for the following questns...

mark 1. - stmt 1 is sufficient.2- stmt 2 is suff.3-both are reqd to solve the questn.4-either is suff.5-both insufficient.

10. what is the hcf of 5 nos., a,b,c,d,e?

stmt 1 - a=72,b=4,c=6

stmt 2 - d= 8, e = 27.

easy set...hope many wud get all correct...

regardsmaxximus

#120 Solution

Answers for yesterday's questions...

1.2222...250 times and 8888...300 times----22..50 times, 88..1500times.2. 333....120 times and 1111...400 times---11..40 times,

33..1200 times.

3. 111...700 times and 9999...200 times.----11...100 times, 99..1400 times.

4. HCF of 33333...200 times. and 777777.....300 times --- 11..100 times, lcm is diff to find... (33333...200 times. * 777777.....300

times) / 11...100 times.

5. 32^250 -1 & 16 ^ 100 - 1----2^50 - 1, 2^10,000 - 1 (nobody got this right)

6. 81^100 -1 & 243 ^ 200 - 1.---- 3^200-1, 3^2000 - 1

7. 343^150-1 & 2401^100 - 1.----7^50 - 1, 7^1800 - 1

8. 125^200 - 1 & 625^120 - 1.---- 5^120-1, 5^2400 - 1

9. 169^320 - 1 & 32^160 - 1.---- 1, prod of 2 nos.

for the following questns...

mark 1. - stmt 1 is sufficient.2- stmt 2 is suff.3-both are reqd to solve the questn.4-either is suff.5-both insufficient.

10. what is the hcf of 5 nos., a,b,c,d,e?


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stmt 1 - a=72,b=4,c=6

stmt 2 - d= 8, e = 27.

answer - option 2. stmt is suffct.This was the only tricky question in the set...look at statement 2...8,27 are co-prime nos. so their hcf is 1. hence, hcf of the entire set is

1. so, stmt 2 is sufficient to answer.

regards

maxximus

#129Quote:

Originally Posted by Rockeeze

hiwas going thru the concept of lcm n hcftype #2

222.........30 times333..............70 times

the method is fine, but cud nt find a logic for taking the lcm of 30 n 70

22......... 30 times=2*111..........30 times

n 333......70 times=3*111.........70 times

cud u plz explain the logic??

111...30 times can be written as

111...10 times x 10^20 + 111....10 times x 10 ^10 + 111...10 times.

hence, v can say 111...30 times is divisible by 11...10 times.

similarly..

111...70 times can be written as

111...10 times x 10^60 + 111...10 times x 10^50 +......111...10 times.

hence this is also divi. by 111...10 times.

so, the hcf is 111...10 times.

please note that 10 times is the max no. of times 11..can be written so that it divides both 30 times n 70 times...

while finding lcm, v need 11... as many times that 11...70 times as well as 11..30 times...so it shud be 111....210

times.


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do revert..

regards

maxximus

#127Quote:

Originally Posted by Rockeeze

ori ginall y posted by iyervani 30.05.07

One of the smaller sides of a right angled triangle is (2^2)*(3^3)*(4^4)*(5^5)*(6^6)*(7^7) . It is known thatother two sides are integers.How many triangles of this type are possible.

The approach mentioned was:

the given number is 2^16 * 3^9 * 5^5 *7^7

a^2 = c^2 - b^2 = c-b * c+bboth c-b and c+b be either odd or even ..but here both cannt be odd..so both even

to wr ite as a product of 2 numbers we can use all powers of 2 excepth the 32th power..so no of tr iangles = (31*19* 11*15 -1 )/2

I did not understand the highlighted part...

could someone plz explain this or any other approach to solve the ques.????

well i thought dis ques was worth a discussion in this thread

vanis query was also not ans der..

i too hv the same query

i think the ans should be 17*10*6*8-1/2

number of ways the samller side cn be expressed as a product of two no

hi rock...thanx for bringing this to the thread...

this question has to do with a concept i've already discussed on this thread...concept of factors...yeah...this

question has lotsa twists n turns attached.

the final answer given by whoever answered it is correct but the xplanation has more than one mistake...seems

the question was solved after seeing the correct answer...another mal-practice that must be avoided...lets have

an elaborate discussion...

the given number is 2^16 * 3^9 * 5^5 *7^7

a^2 = c^2 - b^2 = c-b * c+b

for convenience, take c-b = t, c + b = u.


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both t and u shud be either odd or even ..but here both cannt be odd (since a pythagorian triplet withhypotenuse = even n rest 2 sides = odd does not exist)

so both even

now a^2 = 2^32 * 3^18 * 5^10 *7^14.

now, since both t & u are even, 2x2 already exists in t * u. so the powers of 2 which can be floated across 2

terms to be multiplies reduces by 2. i.e. 32 -2 = 30.

now, the no. of ways in which pairs can be formed are 31 * 19 * 11 * 15. but lets not forget out of these cases,

there lies a case when t = u....or lets say b = 0.

that particular case shud be discarded...

since the factors wud be used in pairs, answr shud be half the no. of factors...so the final answer is...

(31*19*11*15 -1 )/2

please feel free to revert..

with due regards

maxximus


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#132

Today's concept : To solve last digit problems.

Type # 1:

questions of the form...

2003x 2004 x 4235161006 x 432657178001 x 42315098002 x 423087004

concept:

Last n digits of any product depends on the product of last n digits. so just multiply last n digits of each term...find the

product, take last n digits of the product n multiply it with the next term...continue this for all terms.

suppose v had to find last digit of the product above...

so multiply last digits.

3 x 4 = 12. dont worry abt 1 in 12. just remember 2 and multiply it with next no. 2 x 6 = 12. so 2, 2 x 1= 2, 2 x 2 = 4, 4 x 4 = 16.

so, the ans is 6.

for last 2 digits:

03 x 04 x 06 x 01 x 02 x 04 = 76

for last 3 digits:

003 x 004 x 006 x 001 x 002 x 004 = 576.

trust me, last 4 digits wont be asked...as then it becomes bulky...questions in cat are tricky...

Type # 2:

of the form...last digits of 432^43567. i.e base ^ power

i know most of us know this concept...wud take it concisely...

look for the variation in last digit of higher powers of last digit of the base...i.e. 2 here.

2^1 = >22^2 => 42^3 => 82^4 => 62^5 => 2

so we can say that 2,4,8,6 will keep repeating...no. of different digits that the last digits of higher powers can take is known as

cyclicity. every digit has a cyclicity.

to find last digit, find last digit of :

(last digit of base) ^ (power % cyclicity of last digit)


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when, power % cyclicity of last digit = 0,

take,

(last digit of base) ^ (cyclicity of last digit)

in the above example...

432^43567.

2^ (43567 % 4) = 2^3 = 8 answer.

with little practice....u can do all such questions mentally.

Type # 3

Find last 2 digits of 34291^201.

when the last digit is 1,9,0,5...try finding a pattern in last 2 digits...u'll get one...then solve the question accordingly...

last 2 digits of 91^1= 91last 2 digits of 91^2 = 81last 2 digits of 91^3 = 71last 2 digits of 91^4 = 61.......last 2 digits of 91^10 = 01last 2 digits of 91^11= 91

see, we again got 91 as last 2 digits...so v can say that the cyclicity of 91 for last 2 digits is 11 -1= 10

so the answer shud be 91.

lets take one more example...

find last 2 digits of(49)^(37)^(38 )^(39)...(3700)

see...

last 2 digits of 49^1 => 49last 2 digits of 49^2 => 01last 2 digits of 49^3 => 49last 2 digits of 49^4 => 01.

can v say that for all odd powers, answer wud be 01?


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yes!

since the power of 49 is odd...the answer shud be 01.

what if the question was

(49)^(37)^(38 )^(39)...(3700)%100 = ?answer wud be 01%100 = 1

(49)^(37)^(38 )^(39)...(3700) % 20 = ?answer wus be 01 % 20 = 1.

bcoz, to c remainder with 20, v need last 2 digits only.

(49)^(37 )^(38 )^(39)...(3700) % 10

=> 49 % 10 = 9

this method might become tedious when the last 2 digits are unfriendly...like 37, 82 etc. but i have neva seen

such figures appearing in cat...to solve such figures, v need euler's or binomial...ill be taking it when v discussremainders...today's questions are based on the above concept only..

Questions for today..

1. last digit of 3677^400 - 689^84

2. last digit of 11^11 + 12^12....1000^1000.

3. Last 2 digits of : 233527 x 54725 x 64535379 x 64536247 x 63546342 x 435272599 x 7454453279 x

63546337 x 543624334547 x 74547459 x 7454373 x 6545347359 x 735473451...do i need to say....its

tricky...not lengthy!!

4. no. of zeroes at the end in 1^1x2^2x3^3x...250^250.

5. no. of zeroes at the end in 1! x 2! x 3! x...25!

6. Last non-zero digit of 25!

7. Last non-zero digit of 1! x 2! x 3!...15!

Find last 2 digits of...

8. 81^(371)^(372)^...(400)

9. 11 (25)^(31)^(41)...(1001)

10. 7 ^ 2501.

11. 3^2537837.


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Hope the post helps...

with warm regardsmaxximus

#137hey maximus,thnks for such great posts..

cud u just explain the above type 3 again..

thanks,rushi

[/quote]

hi rushi....welcome to the thread...thanx for the interest u've shown here...

wud've been gr8 if u cud cum up with a specific doubt instead of a mere 'please repeat'... i think i was quite

lucid....if u can read the post 2-3 times, am sure u'll understand...or will atleast cum up with a specific

doubt...anyway...i'll try my best again...

suppose v have a question... find last 2 digits of 2001^ 20010.

see last 2 digits of base...n try finding a pattern in last 2 digits of its higher powers..

last 2 digits of 2001 => 01

o1^1 = 01

01^2 = 01

01^3 = 01

so no matter wat is the power of last digits are bound to be 01. hence answer is 01.

lets takle one more example.

23252399 ^ 54635372528263

last 2 digits of base....99...so lets try finding a pattern with 99.

99 ^ 1 = 99

99^2 = 9801

99^3's last 2 digits shud be 01 x 99 = 99


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99^4 => 01

cn v say that wheneva power is odd, answer is 99...when its even, answer is 01.

since the power is odd...answer shud be 99.

hope its clear now...do feel free to revert...lemme know if its still murky..

about the next part...dont be confused...if u know last 2 digits of a no. i can find remainders with 100, 50 , 10,

20 , 25, by merely dividing last digits with the divisor...think for a while over this...u'll understand..

please read this as well as original post 2-3 times, the concept wud be cemented...in cat, we generally find

questions which are more tricky n less formula based...the examples taken here, as well as questions in today's

post form a gud set of tricky questions...

a sincere request to everybody...kindly read the posts more than one time when u aint able to undertsand a

concept...so that u know in which specific step u have a query...else i wud be unnecessarily elaborating over

concept awready clear...if the entire process is not clear, plz mention it...i'd tell sum other approach...but plz be

clear in the posts as to where exactly the doubt lies...that wud save a lot of time n space...

@rockeezee....u're absolutely right bro!!!..dont have to actually take an ex. if two nos are both even or both

odd...their sum as well as difference shud be even...just an observation...

@dranzer...many wrong answers bro...plz check..

@everybody...today's set of questions is tricky...try it!!

@rushi, everybody...thanx look good only below the posts that u appreciate...i dont think v need to pay personal

thanx in a post meant for quiries...let thx be jus official...wat say guys??...

regardsmaxximus

#135 Solution #132hie....answers...!!

1. last digit of 3677^400 - 689^84---- 0

2. last digit of 11^11 + 12^12....1000^1000.----nobody got this right

3. Last 2 digits of : 233527 x 54725 x 64535379 x 64536247 x 63546342 x 435272599 x 7454453279 x


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try these questions...discuss them on the thread...few of them are a bit lengthy but very useful from practice / concept point of

view...i've deliberately made them long...to let everybody understand wat happens to the last digit when a no. is multiplied with

15,20,25 etc. take ur time to solve...as i said...this set is v useful and wud bring in lotsa new concept as u solve it..

with warm regards

maxximus[/quote]

#344

-----------------------------------------------------------

Quantitative Question # 49

------------------------------------------------------------

Question:

Twinkle tells Raveena that she has got 3 kids and 2 of these kids are twins, and also that theirages are all integers. She tells Raveena the sum of the ages of her kids and also the product of

their ages. Raveena says that she has insuficient information to determine the ages, but one

possibility is that the twins are a prime number of years old. If Twinkles twins are teenagers

and their age is not prime, then the sum of the ages of her kids is

(a) a prime number

(b) is greater than 43

(c) a composite number

(d) exactly 2 of the foregoing

(e) still undeterminable

Hi puys,

Its been many days since I last posted on PG, but I was occupied with unavoidable commitments , and now it

seems, I have lot of catching up to do...

Firstly solution to QQAD # 49 (Sorry for the delayed post)

As its mentioned in the question that the age of the twins is a teen non-prime no. hence it can be either 14, 15,

16 or 18....

Probable AgesSumProductAges Raveena Thought

14, 14, X 28 + X 196X 7,7, 4X (not satisfy add. criteria)

16, 16, X 32 + X 256X 2,2, 64X (___ DO____)

18, 18, X 36 + X 324X 2,2, 81X (___ DO____)


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...............................................3,3 , 36X (___ DO____)

15, 15, X 30 + X 225X 3,3, 25X (satisfies all criteria for X=1)

Thats why the ages are 15, 15, 1....

I think i am clear enough :smile:

Regards,

Mohit

PS - There have been some doubts regarding this question. Kindly read the conversation between me and

vineetvijay, from here onwards..... it will help, grasping the problem..

#345

Originally Posted byjunoonmba

Solve this...really nice

There are 1000 soliders with a no. on their T Shirts (1,2,3.....999,1000) standing on a circular track.

Man with no.1 is carrying a sword in his hand. He kills man at no. 2 and pass on sword to no. 3.This goes on until we have only one solider on the track.

What will be the no. on his T-Shirt?

If n is the last number on the T-shirt (or n is the total number of people), then we observe the following pattern

n___________________last surviour T-shirt number

1___________________12/3_________________1/3

4/5/6/7______________1/3/5/7

2^n/..... /2^(n+1)-1____1/3......./2^(n+1)-1

so for n=1023, the surviour will be 1023. and from back tracking from 1022 to 1000, we get for n=1000, thesurviour will be 1023-2*23 = 1023-46 = 977

Hope thats correct!!

Regards,

Mohit
http://www.pagalguy.com/forum/quantitative-questions-and-answers/23536-concepts-total-fundas-post795917.html


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Tip: For these kind of questions, where you can find the answer by mechanical calculations, but nis very

large, try to find a relation in the answer and n, or in other words try to generalize the problem for any

value of nand substitute the original large nin the generalized relation to get the answer..... key thing is

to realize that in which question, generalization will work...


--Unknown

#192wud request everybody to pour in their self-discovered / short cut methods....

fi nd remainder when divided by 100...

1. 1^1 + 11^ 11 + 111^111 +.... thousand ones thousand ones ----90

The remainder when divided by 100, depends upon the last two digi ts...

So the last two digits of the expression are

01 + 11 + 11 +....... = 01 + 999*11(last two digi ts of i t ) = 01+89 = 90

2. 2^20,000----76

2^20000=(2^ 4)^5000=16^5000

Now the cyclini ty of 16 for the last two digi ts is 5....

So the answer is 16^5= 76

3. 1^ 1 + 21 21 + 31^ 31 + ....1001^1001----90 (thanx for noticing missing 11^ 11...ya i know i have a history

of typo...sti ll ...as rock said...i ts always safe to solve wats prin ted!! )

last two digits of the expression i s given by

(01 + 21+ ..... +91 )+ (01+11 + 21 +...... +91 .... 10 times(ti ll 991) + 01 (for 1001)


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449 + 460 *9 + 01 = 449 + 4140 + 01 = 4590

so the answer i s 90

4. 5^ 5 + 15^15 + 25 25 +...125^ 125----25

every expression gives 25 as last two digi ts

25 + 25 +...... 13 times....= 25*13=25

5. 9^ 9 + 99^99 +.....hundred 9s hundred 9s----90

89 + 99*99 = 89 + 01 = 90

kudos to MOHIT, ROCK, SAURABH, SUMI, JHA for their correct answers as well as for the healthy

discussions....keep up the spirit guys!!!!

regards

maximus

#193Today's concept : Mastering Data suffieciency.

With no. of options for every question increased to 5, the expected answer set for D.S. is...

1. Statement 1 alone is sufficient

2. statement 2 alone is sufficient.

3. both statements together can answer the question but none of them is alone sufficient.

4. either statement is sufficient.

5. both statements are insufficient.

wud like to share my approach towards D.S.

first n the mother of all other ...see if statements 1 is sufficient.

if yes...answer shud be 1 or 4.


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............now check if statement 2 is sufficient.

..................if yes, option 4 is correct.

...........................if no, option 1 is correct. dont see if both together can answer the que.

if no....answer can be 2,3 or 5

...........check if stmt 2 is sufficient.

.................if yes, option 2 is correct. dont see if both together can answer the que.

..................if no, answer shud be 3 or 5.

.......................check if both 1 & 2 together lead to a unique solution.

.............................if yes, option 3 is correct

....................................if no, option 5 is correct.

do not forget...

1). The answer shud be unique...even 2 cases suggest that the statement is not suffiecient.

2). The answer shud be consistent...if the answer is no...it shud be always no...if yes...always yes.

3). Both statement can give different answers...they're independent till used together...so even if the answers r

different...the answer wud optn 4 if each statement satisfies above 2 conditions independently.

4). at times u'll see that question can be answered using any one option as well as both options...in such a case,

answer wud be 1 or 2 and not 3.

5)the most important rule....everytime u're able to sort a D.S. question easily...check twice...there must

be a catch sumwhere...

Questions for today...

1. Statement 1 alone is sufficient

2. statement 2 alone is sufficient.

3. both statements together can answer the question but none of them is alone sufficient.

4. either statement is sufficient.

5. both statements are insufficient.

1. is X positive?

stmt 1. x^2 - 5x + 10 = 7

stmt 2. sqrt (x) is real.


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2. is X even?

stmt 1. 7x + 5y = even

stmt 2. 3x + 38y = odd.

3. three packages combinedly weight 60 kg. what is the weight of heaviest package ?

stmt 1. one package weighs 10 kgstmt 2. one package weighs 32kg

4. is X > Y?

stmt 1. 2/5 X > 3/8 Y

stmt 2. 3/7X > 15/29 Y

5. If p^q = 289, (p-q)^(q/2) = ?

stmt 1. p > q

stmt 2. p is not = 1.

6. for above questn.

stmt 1. p>q

stmt 2. p is prime.

7. are sum goats not cows?

stmt 1. all cows are lions

stmt 2. all lions are goats.

8. is the no. of flowers in the garden = 3576876767576654544334538?


st. 2 no. of rows = no. of columns.

9. for above question


st 2. no. of flowers are even.

10. x = ?

st. 1 x^2 - 5x + 6 = 0


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st. 2 x^3 - 8 = 0

Happy solving!!!

regards

maxximus

#Solution 193

1.The statement 2 alone is sufficient. hence the answer is 2

2. statement 1 alone is not sufficient and but statement 2 alone can be, so the answer is 2

(Credits to rockezee for suggesting correction)

one more thing maximmus, in the question, we have to just na that whether we can find out the condition ornot... as in, if we can find using any of the 2 statements (or both of them), that x is odd, we can answer the

question "is X even?"clarify please..

3. statement 1 alone is not sufficient, but statement 2 alone is sufficient (32>60's 50 %), hence the answer is 2

4. statement 1 alone is sufficient and statement 2 alone is also sufficient, hence answer is 4

5.statement 1 alone is not suffcient and statement 2 alone is also not sufficent.and even 1 and 2 both do not leadto imply a unique result .Hence answer is 5

6.statement 1 alone as not suffcient and statement 2 alone is also not sufficent.and still 1 and 2 both do not lead

to imply a unique result (p being 17 and q being 2 or p being 289 ^ 4 and q being 0.25).Hence answer is 5.

7. These kinda question would be easy with the help of venn-diagrams, I suppose. But still, as per max'smethod.....

statement 1 alone is not sufficient and statement 2 is also not sufficient alone and both of them together also do

not lead to a conclusion. hence the answer is

8. statement 1 alone is not sufficient and statement 2 alone is not sufficient. even 1 and 2 together do not lead toa unique solution.... Hence the answer is 5

9. Same explanation as given in 8. hence the answer is 5

10. statement 1 alone is not sufficient. but statement 2 alone is sufficent. So the answer is 2

As i am the first to answer , I have taken the liberty to quote the questions.....


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--Unknown


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1)Let S be the set of five-digit numbers formed by the digits 1, 2, 3, 4 and 5,

using each digit exactly once such that exactly two odd positions are occupied

by odd digits. What is the sum of the digits in the rightmost position

of the numbers in S?

a. 228 b. 216 c. 294 d. 192

2)There is a circle of radius 1 cm. Each member of a sequence of regularpolygon P1(n), n = 5, 6...., Where n is the number of sides of polygon is

circumscribing the circle and each member of a sequence of regular polygon

P2(n), n = 4, 5.. where n is the number of sides of a polygon, is incribed

in the circle. Let L1(n) & L2(n) denote the perimeter of the corresponding

polygons P1(n)& P2(n). let X=(L1(13)+2pie)/L2(17) Then

a)pie/4 < X < 1 b)1 < X < 2 c)X > 2 d)X < 2


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#408I have analysed the concept and If my understanding is right then I can modify the concept as.......[Please

Verify....]

Concept-1:-

(N^p-1)%p gives remainder as 1(p-1 is in power)

Nmenoic is Remainder will be 1 [and Do all opearations on 1.............]

Concept-2:-

(p-1)!%p leaves remainder p-1

Nmenoic is Remainder will be Numerator

Now why I modified is because it will remove the burden to learn where to add 1 or subtract 1

Now we can easily do the calculation in a moment...........

Suppose question is ....

(91^28-24)%29 leaves remainder

Now we Know 91^28%29 leaves remainder 1

therefore 91^28-24%29 will leave remainder 1-24i.e -23i.e 6...............

And

(198!+177)%199

then we can say 198!%199 leaves reamainder 198 .Therefore 198+177=375%199 leaves remainder -23 i.e

176..........

Correct me If i Am Wrong...............

Concepts for CATH

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