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Maizi from Norwich asks, "Four positive whole numbers add up to 84 One of the numbers is a multiple of 17 The other 3 numbers are equal. What are they?"

Hi Maizi, as is my custom with math problems, I'm not going to answer your question; instead I'm going to invent a similar problem and show you the technique for solving it. This will allow you the satisfaction of solving this particular problem on your own. So here is my problem:

**Five positive whole numbers add up to 97, and one of them is a multiple of 19. The other four numbers are all equal. What are they?**

First, we consider that if we add together four equal numbers, the result must be a multiple of four. For example, 7 + 7 + 7 + 7 = 28, or 11 + 11 + 11 + 11 = 44.

Algebraically, we'd say that n + n + n + n = 4n.

So this means that if we subtract the first number (the one that's a multiple of 19) from 97, we will have a result that is a multiple of 4.

What could the first number be? 19, 38, 57, 76, or 95. Any other value is larger than the sum of 7, which is not possible if all the integers are non-negative.

Based on this, we can find all possible sums of the other four numbers, by subtracting each of the numbers above from 97:

97 - 19 = 78

97 - 38 = 59

97 - 57 = 40

97 - 76 = 21

97 - 95 = 2

Remember what I said earlier? The other four numbers must add to a multiple of 4. The only one of those sums above that is a multiple of 4 is 40. Therefore, we conclude that the first number is 57, and the other four numbers are each 10 (because 40 / 4 = 10).

I hope that's helpful; good luck with your version of the problem!