1987 AMC 8 Problem 8

Attempt Problem 8 of the 1987 AMC 8 below, then check your answer against the professionally curated solution 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).

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
Answer: B
Concepts:place valuebounding to limit cases
Small hint:
The number 98769876 is already close to 10,000,10{,}000, so adding anything pushes the sum past it
Big hint:
Check the largest possible sum too, using A=B=9A = B = 9
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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