# Divisible By 5

## Introducing the Problem

The problem is,what is the smallest number divisible by 5 but leaves a remainder of 1 when divided by 2, 3, or 4 is 25. What is the second smallest number that leaves a remainder of 1 when divided by 2, 3, or 4?

## Explanation

The least common multiple(LCM) of 2, 3, and 4 is 12 but if you add one because there is a remainder then it will be 13. 13 isn't divisible by 5 so it isn't the smallest multiple. Now do you get it? LCM means the least common multiple, like you can find it by finding all the common multiples and the on that is the least is the least common multiple.

## The Method

The method we will be using in this problem is guess and check. Guess and check is one of the most powerful method of solving problems. Also it will come in use for this problem because this is the guess and check type of problem. This isn't all guess and check but most of it is. So the method we will be using is guess and check sort of.

## Step 1

The first thing we will need to do is to find the LCM of the three numbers 2, 3, and 4. The LCM has to end in a four because there is no way it can end in a 9 so it has to end in a 4. The reason behind this is because 5's divisiblility rule is it has to end in a 5 or a 0. And also because the remainder of 1 you will have to subtract both numbers by. Therefore means you need to have a 4 or a 9.

## Step 2

After you found your LCM (which should be 12) you add it by 1. If the one digit is 5 or 0, then that is you final answer. If it isn't, then multiply your LCM by 2 then 3 then 4 and so on. Keep going until you have a number that ends in 4. The first number should be 24 but that is the one in the example. You have to find the second number. The reason behinds this step is that you have to find the second number that leaves a remainder of 1 when divided by 2, 3, and 4. And it has to be divisible by 5 so this method is most likely the most useful method. So the answer should be 85. If you find the second number early then your most likely wrong.

## Conclusion

So all in all, this is a guess and check sort of problem. This problem isn't all guess and check but most of it is.