1997 AMC 8 Problem 5

Below is the professionally curated solution for Problem 5 of the 1997 AMC 8, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1997 AMC 8 solutions, or check the answer key.

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Concepts:multiple digits systematic listing

Difficulty rating: 730

There are many two-digit multiples of $7,$ but only two of the multiples have a digit sum of $10.$ The sum of these two multiples of $7$ is

$119$

$126$

$140$

$175$

$189$

Solution:

Listing out all the two-digit multiples of $7,$ we get $14, 21, 28, 35, 42, 49, 56, 63, 70,$ $77, 84, 91, \text{ and } 98.$ We have that $28$ and $91$ are the two multiples whose digits add to $10.$

The sum of these numbers is $28 + 91 = 119.$

Thus, A is the correct answer.

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