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Well, we know that the sum is at most $14+13+12+11+10+9+8+7=84$, so the product is at most $7056$.

If there are $7$ or more children, then the product is at least $8!>7056$, so there are at most $6$ children.

Furthermore, if there are $6$ children, the sum is at most $84-8-7=69$, so the product is at most $69^2=4761$, but the product is at least $7!=5040$, so there cannot be $6$ children.

Let $S$ denote the sum and $P$ the product. By the AM-GM inequality, we have $\frac{S}{n} \ge \sqrt[n]{P} = \sqrt[n]{S^2}$, so $\frac{S^n}{n^n} \ge S^2$, or $S^{n-2} \ge n^n$, where $n$ is the number of children. This means that $n \ge 3$, since $n^n \ge 1$, which would contradict $n=2$.

I'm not sure there's much else you can do without getting into some messy casework. To rule out $3$ and $5$, you can divide into cases like "Suppose at least two children are older than 11," etc, and use similar arguments regarding sums and products as above. To then find the result given that $n=4$, you'll need to use some divisibility arguments and, yes, a little bit of casework.

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Well, we know that the sum is at most $14+13+12+11+10+9+8+7=84$, so the product is at most $7056$.

If there are $7$ or more children, then the product is at least $8!>7056$, so there are at most $6$ children.

Furthermore, if there are $6$ children, the sum is at most $84-8-7=69$, so the product is at most $69^2=4761$, but the product is at least $7!=5040$, so there cannot be $6$ children.

Let $S$ denote the sum and $P$ the product. By the AM-GM inequality, we have $\frac{S}{n} \ge \sqrt[n]{P} = \sqrt[n]{S^2}$, so $\frac{S^n}{n^n} \ge S^2$, or $S^{n-2} \ge n^n$, where $n$ is the number of children. This means that $n \ge 3$, since $n^n \ge 1$, which would contradict $n=2$.