Question

What ir the smallest three-digit number which when divided by 6 leaves a remainder of 5 and when divided by 5 leaves a remainder o3 3

Answer 1

First find the smallest 3-digit number that is evenly divisible by both 5 and 6. Then add 1.

Since 5 is prime and not a factor of 6, the smallest positive integer evenly divisible by both 5 and 6 must be 5 times 6 or 30. So the question becomes what is the smallest 3-digit number evenly divisible by 30? 1, 2, and 3 times 30 each produce a 2-digit number. Hence the smallest integer multiplier of 30 that produces a 3-digit product is 4, and the smallest 3-digit integer divisible by 30 is 120. Hence, 120 is the smallest 3-digit integer evenly divisible by both 5 and 6. And finally, 121 is the smallest 3-digit integer when divided by either 5 or 6 leaves a remainder of 1 in each case.