2006 AMC 8 Problem 11 — Solution | LIVE by Po-Shen Loh

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2006 AMC 8 Problem 11

Below is the professionally curated solution for Problem 11 of the 2006 AMC 8, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 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:digits perfect square casework

Difficulty rating: 1310

11.

How many two-digit numbers have digits whose sum is a perfect square?

$13$

$16$

$17$

$18$

$19$

Solution:

There is $1$ number whose digit sum is $1: 10.$

There are $4$ numbers whose digit sum is $4: 13, 22, 31,$ and $40.$

There are $9$ numbers whose digit sum is $9: 18, 27, 36, 45, 54,$ $63, 72, 81,$ and $90.$

There are $3$ numbers whose digit sum is $16: 79, 88,$ and $97.$

Therefore, there are $17$ numbers that satisfy the problem statement.

Thus, C is the correct answer.

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