1993 AMC 8 Problem 20 — Solution | LIVE by Po-Shen Loh

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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 $10^{93} - 93$ is expressed as a single whole number, the sum of the digits is

$10$

$93$

$819$

$826$

$833$

Answer: D

Concepts:place value digits

Difficulty rating: 1140

Solution:

Subtracting $93$ from $10^{93}$ (a $1$ followed by $93$ zeros) gives a number that is $91$ nines followed by $07.$

The digit sum is $91 \times 9 + 0 + 7 = 819 + 7 = 826.$

Thus, the correct answer is D .

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