Find the smallest number that leaves the remainder 13, 41 and 29 on being divided by 20, 48 and 36 respectively.

When divided by 20 remainder is 13. So, if we add 7 and then divide by 13 remainder, remainder will be 0 (as 13+7 becomes 20.

Similarly after adding 7, remainder when divided by 48 becomes 0. Similarly, for 36 it becomes 0.

20 is 4×5

48 is 16x3

36 is 4x9.

The factors are not taken randomly. They are powers of primes.

Common multiple has to be multiple of 16 (highest power of 2); otherwise it will not be multiple of 48.

Similarly, if should be multiple of 9 and 5.

Hence, it should be multiple of 720. Smallest such positive number is 720.

This is divisible of 20, 48 and 36

But this number is after adding 7; so required number is 713.

C is correct choice.