Latest TCS Aptitude Question SOLUTION: Think of a 135 digit number formed by writing the first 72 natural number side by side in a string.(i.e.) 1 2 3………70 71 72. Find the remainder when it is divided by 12. a. 2

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Think of a 135 digit number formed by writing the first 72 natural number side by side in a string.(i.e.) 1 2 3………70 71 72. Find the remainder when it is divided by 12.
a. 2
b. 9
c. 0
d. 5

Asked In TCS Kalyani Reddy (10 years ago)

Unsolved Read Solution (9)

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Read Solution (9) :

anusha  (10 years ago)

ans c
12345678..........7172 is a 135 digit number
divisibility rule for 12 is :divisible by 4 and divisible by 3
for divisible by 3 sum is divisible by 3
for divisible by 4 last two nos should be divisible by 4
sum of 72 nos is 72*73/2=36*73
then 36*73/3 and last two digits 72 is divisible by 4 so remainder is zero

Is this solution Helpfull? Yes (25) | No (1)  

Deepak Kumar Sahu  (10 years ago)

ans 0, because it is divisible by 3 and 4

Is this solution Helpfull? Yes (18) | No (3)  

MUNISH YADAV  (10 years ago)

Sum of the first 72 natural numbers is =72(72+1)/2=2628
2628 is divisible by 3,9 (divisibility rule for 3,9)
123456......7172 is divisible by 4 also as last two digits of this number which is 72 is divisible by 4 (divisibility rule of 4)
Option:B

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MUNISH YADAV  (10 years ago)

Option: C (sorry)

Is this solution Helpfull? Yes (3) | No (5)  

Sukanya Das  (10 years ago)

option:C,bcz sum of 1st 72 natural no is 72*73/2=2628 and if it is divided by 12 dn remainder is 0

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Miriyala. siva sankar  (10 years ago)

c.0 to divisible by 12
last two digits are must divisible by 3 &4
so result 0

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Kalyani Reddy  (10 years ago)

The 135 th digit in 1,2,3,.....72 is 2.
so the number is 72
72/12 remainder is 0.
option:C

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sonali ingale  (10 years ago)

c....12 is divisible by 4 n 3...........so by divisibility rule of 4 n 3...it is satisfied............so remainder is 0

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DHIVYAVARSHINI  (6 years ago)

divisibility rule for 12: the number should divide by both 3 and 4
so taking last two digit 72 it is divisible by 4 and while considering their sum by using the fomula n(n+1)/2, it is found that it is divisible by 3 too
so the remainder is option c) 0

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