1994 AMC 8 Problem 23

Attempt Problem 23 of the 1994 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 1994 AMC 8 solutions, or check the answer key.

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

23.

If X,X, Y,Y, and ZZ are different digits, then the largest possible 33-digit sum for
XXXYX+X\begin{array}{cr} & XXX \\ & YX \\ + & X \\ \hline \end{array}
has the form
XXYXXY
XYZXYZ
YYXYYX
YYZYYZ
ZZYZZY
Answer: D
Concepts:cryptarithmplace value
Small hint:
Write the addition in columns; if XX is too large the hundreds place carries and the answer gains a fourth digit.
Big hint:
Take the largest XX that keeps the sum to three digits, then choose YY to make the total as large as possible and see which letters the resulting digits match.
Solution:
The hundreds digit of the sum comes from XX plus any carry, so if X=9X = 9 the sum would spill over into four digits. The largest allowed value is X=8.X = 8.
To make the sum as large as possible, take Y=9:Y = 9: then 888+98+8=994.888 + 98 + 8 = 994. Its digits are 9,9, 9,9, 4;4; since 9=Y9 = Y and 44 is a new digit Z,Z, the sum has the form YYZ.YYZ.
Thus, the correct answer is D .
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