2026 AMC 10A Problem 10

The list $1, 4, 9, 16, \dots, n^2$ is just $1^2, 2^2, 3^2, \dots, n^2$, so it has $n$ numbers. The plan is to write the median and the mean as formulas in $n$, then set up one equation.

1. The median. The squares are already in order from smallest to largest, and $n$ is odd, so there is exactly one middle entry. Try a small case first: for $n = 5$ the list is $1, 4, 9, 16, 25$, and the middle entry is $9 = 3^2$. The $3$ is the middle of $1, 2, 3, 4, 5$, which is $\frac{5+1}{2}$. In general the middle of $1, 2, \dots, n$ is $\frac{n+1}{2}$, so
$$\text{median} = \left(\frac{n+1}{2}\right)^2 = \frac{(n+1)^2}{4}.$$

2. The sum. There is a standard formula for adding up the first $n$ squares:
$$1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}.$$
Test it on $n = 3$: $1 + 4 + 9 = 14$, and $\frac{3 \cdot 4 \cdot 7}{6} = \frac{84}{6} = 14$. This formula comes up often on the AMC, so it is worth memorizing.

3. The mean. The mean is the sum divided by how many numbers there are, which is $n$. The $n$ on top cancels:
$$\text{mean} = \frac{(n+1)(2n+1)}{6}.$$

4. Turn the words into an equation. "$32\%$ greater than the median" means the median plus $32\%$ of the median, which is $1.32$ times the median:
$$\frac{(n+1)(2n+1)}{6} = 1.32 \cdot \frac{(n+1)^2}{4}.$$

5. Cancel what you can. Both sides contain a factor $n + 1$, and it is not zero, so divide both sides by it. Also, $\frac{1.32}{4} = 0.33$:
$$\frac{2n+1}{6} = 0.33\,(n+1).$$
Multiply both sides by $6$:
$$2n + 1 = 1.98\,(n+1) = 1.98n + 1.98.$$

1. Solve. Subtract $1.98n$ and $1$ from both sides: $0.02n = 0.98$, so $n = 49$.

1. Check. For $n = 49$ the mean is $\frac{50 \cdot 99}{6} = 825$ and the median is $25^2 = 625$. Since $32\%$ of $625$ is $200$, and $625 + 200 = 825$, the mean really is exactly $32\%$ greater than the median.

The answer is $\boxed{\textbf{(C)}\ 49}$.

When a list follows a clear pattern, its mean and its median can each be written as a formula in $n$, and that turns a statistics question into an ordinary equation. Here the mean is bigger than the median because the large squares at the end pull the average up. The ratio of mean to median works out to $\frac{2(2n+1)}{3(n+1)}$, which creeps toward $\frac{4}{3}$ as $n$ grows, so every answer choice gives a ratio between $1.30$ and $1.33$. That is why the exact equation, not a rough estimate, is needed.

Test the answer choices, using the formulas from Steps 1 to 3.

For $n = 19$ the mean is only $30\%$ greater, which is too small. For $n = 49$ it is exactly $32\%$ greater, so $49$ is the answer. (The ratio keeps growing slowly as $n$ gets larger, so the bigger choices overshoot.)