1987 AMC 8 Problem 8

Below is the professionally curated solution for Problem 8 of the 1987 AMC 8, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1987 AMC 8 solutions, or check the answer key.

Difficulty rating: 930

In the addition problem below, $A$ and $B$ are nonzero digits:

$\begin{array}{r} 9876 \\ A32 \\ +B1 \\ \hline \end{array}$

How many digits (not necessarily different) are in the sum of the three whole numbers?

$4$

$5$

$6$

$9$

depends on the values of $A$ and $B$

Since $A$ and $B$ are at least $1,$ the sum is at least $9876 + 132 + 11 = 10{,}019,$ which has $5$ digits.

At the other extreme, with $A = B = 9$ the sum is $9876 + 932 + 91 = 10{,}899,$ still $5$ digits. So the sum always has exactly $5$ digits.

Thus, the correct answer is B .