What is the smallest positive number which, when divided by 2, 3, 4, 5, and 6, always leaves a remainder of 1?

The key to solving this classic number puzzle is to first look for a number that is one less than the target. If a number leaves a remainder of 1 after being divided by a set of divisors, it means the number immediately preceding it must be perfectly divisible by all of them. So, the real task is to find the smallest positive number that is a perfect multiple of 2, 3, 4, 5, and 6.

This value is known in mathematics as the Least Common Multiple, or LCM. It is the smallest number that all of the divisors can divide into without leaving any remainder. To find the LCM of this set, we can see that any multiple of 6 already handles the 2 and 3. We then need a number that is also a multiple of 4 and 5. The smallest number that fits all these criteria is 60.

Since 60 is the smallest number that 2, 3, 4, 5, and 6 can all divide into evenly, we simply need to add 1 back to satisfy the condition of the original riddle. This brings us to 61. This type of problem is a simple example of modular arithmetic, a branch of mathematics that deals with remainders and cyclical patterns, and is foundational to fields like cryptography and computer science.