1997 AMC 8 Problem 5

Attempt Problem 5 of the 1997 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 1997 AMC 8 solutions, or check the answer key.

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

5.

There are many two-digit multiples of 7,7, but only two of the multiples have a digit sum of 10.10. The sum of these two multiples of 77 is
119119
126126
140140
175175
189189
Answer: A
Concepts:multipledigitssystematic listing
Small hint:
Keep only the multiples whose digits add to 1010
Big hint:
List the two-digit multiples of 77
Solution:
Listing out all the two-digit multiples of 7,7, we get 14,21,28,35,42,49,56,63,70, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77,84,91,98. 77, 84, 91, 98. We have that 2828 and 9191 are the two multiples whose digits add to 10.10.
The sum of these numbers is 28+91=119.28 + 91 = 119.
Thus, A is the correct answer.
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