1994 AMC 8 Problem 23 — Solution | LIVE by Po-Shen Loh

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1994 AMC 8 Problem 23

Below is the professionally curated solution for Problem 23 of the 1994 AMC 8, 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 of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:cryptarithm place value

Difficulty rating: 1090

23.

If $X, Y,$ and $Z$ are different digits, then the largest possible $3$ -digit sum for

$\begin{array}{cr} & XXX \\ & YX \\ + & X \\ \hline \end{array}$

has the form

$XXY$

$XYZ$

$YYX$

$YYZ$

$ZZY$

Solution:

The hundreds digit of the sum comes from $X$ plus any carry, so if $X = 9$ the sum would spill over into four digits. The largest allowed value is $X = 8.$

To make the sum as large as possible, take $Y = 9:$ then $888 + 98 + 8 = 994.$ Its digits are $9, 9, 4;$ since $9 = Y$ and $4$ is a new digit $Z,$ the sum has the form $YYZ.$

Thus, the correct answer is D .

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