When the positive integer "n" is divided by 3, the remainder is 2 and when "n" is divided by 5, the remainder is 1. What is the least possible value of "n"? The answer is 11. My question is, if you have very large numbers to work with in this problem, what would be the quickest and easiest way to calculate this quickly? Thank you for your time and attention to my question.

Chinese remainder theorem.

Basic solution.

LCM(3,5)=15

If we divide by 3 and the remainder is 2, the possible remainders when dividing by 15 are 2, 5, 8, 11, and 14.

If we divide by 5 and the remainder is 1, the possible remainders when dividing by 15 are 1, 6,and 11.

So all solutions are 15n+11 and 11 is the smallest.

Chinese remainder theorem solution.

Again LCM(3,5)=15.

The smallest multiple of 3 and remainder 1 when dividing by 5 is A=6 and the smallest multiple of 5 with remainder 1 when dividing by 3 is B=10.

Dividing A+2B=26 by 15 gives remainder 11.

So again all solutions are 15n+11 and 11 is the smallest.

## Comments

Thank you, Roman. This is a GREAT answer. Thank you so much!! Linda