2002 AMC 10B Problem 25

Let the original list have $n$ integers with mean $m$, so its sum is $mn$. The individual entries never matter, only this sum.

Append $15$: the new list has $n + 1$ entries, sum $mn + 15$, and mean $m + 2$:
$$mn + 15 = (m + 2)(n + 1) = mn + m + 2n + 2 \quad\Longrightarrow\quad m + 2n = 13.$$

Append $1$ to that list: $n + 2$ entries, sum $mn + 16$, and the mean drops by $1$ from $m + 2$ to $m + 1$:
$$mn + 16 = (m + 1)(n + 2) = mn + 2m + n + 2 \quad\Longrightarrow\quad 2m + n = 14.$$

From the first equation $m = 13 - 2n$; substituting, $26 - 4n + n = 14$, so $3n = 12$ and $n = 4$ (with $m = 5$).

Sanity check with four $5$s: adding $15$ gives mean $\frac{35}{5} = 7$ (up $2$); adding $1$ gives $\frac{36}{6} = 6$ (down $1$).

The answer is $\boxed{\textbf{(A)}\ 4}$.

A statement about a mean is a statement about a sum: mean $\times$ count. Parametrizing the list by $(n, m)$ instead of by its entries turns each "the mean changes by" sentence into one linear equation, because the $mn$ products cancel on both sides. Two sentences, two equations, two unknowns.

Test the choices with a constant list. For $n = 4$, choose four copies of some value $m$; the first condition $\frac{4m + 15}{5} = m + 2$ gives $m = 5$, and then $\frac{36}{6} = 6 = 7 - 1$ confirms the second. Since the answer is unique, (A) is it.

• Comparing the second mean to the original mean instead of to the enlarged list's mean (it should be m + 2 - 1 = m + 1 with n + 2 numbers).
• Expanding $(m+2)(n+1)$ or $(m+1)(n+2)$ incorrectly, for example swapping the $2m$ and $2n$ terms, which yields $n = 5$.