Question

My wife has a collection of seashells. She doesn't remember how many she has. She knows that if she counts by 2 there is 1 left over. If she counts by 3 there is 1 left over. If she counts by 4, 5, or 6 there is still one left over. However, if she counts by 7 there are none left over. What are the fewest number of shells that she has?

Answer 1

The smallest number that meets all the given conditions is 301, which is the smallest multiple of 7 that when divided by 2, 3, 4, 5, and 6, leaves a remainder of 1.

Step-by-step explanation:

The question described is a classic problem in number theory, and it involves finding the smallest number that has a remainder of 1 when divided by 2, 3, 4, 5, and 6, and has no remainder when divided by 7. This problem can be solved using the Chinese Remainder Theorem, but for simplicity and educational purpose, we can find the answer through systematic checking of multiples of 7, since the number we are looking for is a multiple of 7 (7, 14, 21, ...), that also fulfills the other conditions. Upon reaching the number 301, we find that 301 divided by 2 leaves a remainder of 1, divided by 3 also leaves a remainder of 1, the same applies when dividing by 4, 5, and 6. Therefore, the smallest number of seashells the student's wife could have is 301.