Find the remainder when 13571357...upto 1000 digits is divided by 101a)1 b)9 c)14 d)92

Find the remainder when 987987987..upto 123 digits is divided by 1001a)1 b)14 c)427 d)987

Find remainder when 123456123456...upto 600 digits is divided by 999a)41 b)579 c)957 d)997

When 100 is divided by 101 rem is -1 nd 1 when divided by 10000 ..... so we'll make the groupf of 2 by alternately taking + and - and taking the net sum ..... grouping will start from the unit digit . 57 - 13 = 44 * 250 ( 250 is no. of grps ) ..... 44 * 48 = 2112 = -9 = 92 . Similarly for 1001 . When 1000 is divided by 1001 it leaves rem of -1 nd 1 with 1000000 . So take groups of 3 by alternately takin + and - . When 100 is divided by 99 it leaves a rem of 1 . Take the grpf of 2 and add all . Reply Like 1 Like 27 Apr '13 #1 Permalink Report

Protibha Chakravorty@Tviprohey thanks but did not understand why are we taking groups of 2 or 3 ..how is that related ..and in the second ques if i take group of three and do + and - ..wouldnt that be 987 - 987 and in the last ques why dividing by 99 ..in ques its 999... Reply Like 0 Like 27 Apr '13 #2 Permalink Report

Grim Reaper@techsurge5.3kbuddy this is a patternremainderfor 11 sum of even - sum of odd.eg. 1234 /11 rem = 4+2-1-3 =2

for 101 sum of groups of 2 (right most alternating )- left most alternating

eg. 1234/101 remainder = 34-12=22 3412 /101 remainder = 12-34 =-22=79123456 /101 remainder (12+56) -34 = -34 =67@Tvipro

pairing always from right most digits for subtraction

abcdef => ef+ab-cdabcdefgh => gh+bc-ab-ef Reply Like 0 Like 27 Apr '13 #3 Permalink Report

Protibha Chakravorty@Tvipro@techsurgehey ..so u mean to say that for 101 we take groups of two and for 1001 ..groups of three and then subtract alternately ..and for the second ques ..all the 987s will be cancelled except for one as there will be 41 groups ...is that right ? also plz explain for the third question dividing by 999 Reply Like 0 Like 27 Apr '13 #4 Permalink Report

Grim Reaper@techsurge5.3k2nd question 9873rd question 957 by CRTwhats the OA ? Reply Like 0 Like 28 Apr '13 #5 Permalink Report

Grim Reaper@techsurge5.3k@Tvipro to confirm answers if u dont have go to wolframalpha and type Number % divisor to get the remainder in an instant Reply Like 0 Like 28 Apr '13 #6 Permalink Report

Protibha Chakravorty@Tviprothe answers r correct dat u have mentioned ..i jus didnt understand how to go about such ques..wats CRT and OA btw Reply Like 0 Like 28 Apr '13 #7 Permalink Report

Grim Reaper@techsurge5.3kOA = Official AnswersCRT = Chinese Remainder Theorem

here u just needed to check remainder when divided by 3 (factor of 999)and eliminatewhich led us to options b and c (next we checked digit sum by but still b anc were there not able to eliminate)after checking with 27 (factor of 999)

the number 123456..... has digit sum 2100, remainder 2100/27 = 21remainder as in options b and c should also give remainder of 21 when divided by 27

only c satisfy and hence b eliminated Reply Like 0 Like 28 Apr '13 #8 Permalink Report

bharat santani@bharat141019second question 1001 is equal to 7*11*13 any no of the form abcabc will be completely divisible by 1001 and abcabcabcabc i.e 12 digits will be completely divisible and similarly up to 120 digits and then last three digits will be 987 which will be remainder (d option) Reply Like 1 Like 28 Apr '13 #9 Permalink Report

Raj @rajivrajput0079Answers are d) c) c) Reply Like 0 Like 06 May '13 #10 Permalink ReportComments Page 2 of 2 Timeline Most Liked Timeline

ravi prasad@ravi63891.1kCan anybody plz tell me whats wrong in my approach ?? Reply Like 0 Like 21 May '13 #11 Permalink Report

ravi prasad@ravi63891.1kI got the digit sum as 2100 . Now a/c to TIME , Rem (N/999) = Rem(Sn/999) where Sn is the sum of the digits of the number. So when i do 2100 / 999 i get 102 as the remainder, which is not the right answer.