1993 AMC 8 Problem 20

Attempt Problem 20 of the 1993 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 1993 AMC 8 solutions, or check the answer key.

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

20.

When 1093−9310^{93} - 93 is expressed as a single whole number, the sum of the digits is
1010
9393
819819
826826
833833
Answer: D
Concepts:place valuedigits
Small hint:
109310^{93} is a 11 followed by 9393 zeros; subtracting 9393 changes only the right end
Big hint:
The result is a block of nines ending in 07;07; determine how many nines there are
Solution:
Subtracting 9393 from 109310^{93} (a 11 followed by 9393 zeros) gives a number that is 9191 nines followed by 07.07.
The digit sum is 91×9+0+7=819+7=826.91 \times 9 + 0 + 7 = 819 + 7 = 826.
Thus, the correct answer is D .
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