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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:place valuebounding to limit cases

Difficulty rating: 930

8.

In the addition problem below, AA and BB are nonzero digits:

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

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

44

55

66

99

depends on the values of AA and BB

Solution:

Since AA and BB are at least 1,1, the sum is at least 9876+132+11=10,019,9876 + 132 + 11 = 10{,}019, which has 55 digits.

At the other extreme, with A=B=9A = B = 9 the sum is 9876+932+91=10,899,9876 + 932 + 91 = 10{,}899, still 55 digits. So the sum always has exactly 55 digits.

Thus, the correct answer is B .

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