Problem 3. Little kids were eating candies. Each one ate 7 candies less than all the others together, but still more than one candy. How many candies were eaten in total?

[5 points] (A. V. Shapovalov)
Answer: 21 candies.

Solution. Let's choose one of the children, for example, Petya. If we take away 7 candies from all the others, there will be as many left as Petya has. This means that twice the number of candies Petya has equals the total number of candies minus seven. The same can be said about any of the children, which means that all the children have the same number of candies - let's say, in piles.

It is clear that each one ate one whole pile less than the others together. Therefore, 7 is divisible by the size of the pile. Since each one ate more than 1 candy according to the condition, there are 7 candies in each pile, i.e., each one ate one pile less than the others together. Petya ate one pile, so the others ate two. Therefore, there are three piles in total, and 21 candies.

This solution can also be written algebraically.

Let $S$ be the total number of candies eaten by the children. If one of the children ate $a$ candies, then according to the condition, all the others ate $a+7$ candies, and together they ate $S=a+(a+7)=$ $=2a+7$ candies. This reasoning is valid for each child, so all the children ate the same number of candies: $a=(S-7)/2$ each.

Now let $N$ be the number of children. Then the condition can be written as $a=a(N-1)-7$, from which $7=a(N-2)$. The number 7 is prime, so one of the factors is 1, and the other is 7. But according to the condition $a>1$, so $a=7, N-2=1$. Thus, $N=3$, and a total of $S=aN=21$ candies were eaten.