1. **State the problem:** There were some children at a party. Each child took a gift for each of their friends, and altogether there were 81 gifts. We need to find out how many children were at the party. 2. **Define variables:** Let the number of children be $n$. 3. **Analyze the situation:** Each child has $n-1$ friends (because a child doesn't give a gift to themselves). 4. **Calculate total gifts:** Each of the $n$ children took gifts for $n-1$ friends, so the total gifts is $$n \times (n-1) = 81.$$ 5. **Form the equation:** $$n(n-1) = 81.$$ 6. **Rewrite the equation:** $$n^2 - n - 81 = 0.$$ 7. **Use quadratic formula:** For equation $ax^2 + bx + c = 0$, solutions are $$n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$$ Here, $a=1$, $b=-1$, $c=-81$. 8. Calculate discriminant: $$\Delta = (-1)^2 - 4 \times 1 \times (-81) = 1 + 324 = 325.$$ 9. Calculate solutions: $$n = \frac{1 \pm \sqrt{325}}{2}.$$ Approximating $\sqrt{325} \approx 18.03$ 10. Possible values: $$n = \frac{1 + 18.03}{2} \approx 9.515,$$ $$n = \frac{1 - 18.03}{2} \approx -8.515.$$ 11. Number of children must be positive integer, so test integer near 9.515. 12. Check integer $n=9$: $$9 \times (9-1) = 9 \times 8 = 72 \neq 81.$$ Check $n=10$: $$10 \times (10-1) = 10 \times 9 = 90 \neq 81.$$ 13. Check if 81 is a product of two consecutive integers: Factor pairs close to $81$ are $(9,9)$ and $(8,10)$. 14. Since no integer $n$ satisfies $n(n-1)=81$, conclusion: Number of children is not an integer based on 81 gifts. **Answer:** No integer number of children satisfies the problem with 81 gifts exactly. If the problem expects an integer solution, then $n=9$ children (72 gifts) or $n=10$ children (90 gifts) are closest.