2014 AMC 12A Problem 16

Attempt Problem 16 of the 2014 AMC 12A 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 2014 AMC 12A solutions, or check the answer key.

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16.

The product

(8)(888\ldots8),

where the second factor has

k

digits, is an integer whose digits have a sum of

1000.

What is

k?

901

911

919

991

999

Multiply small cases:

8\cdot8=64,

8\cdot88=704,

8\cdot888=7104

The product is

7\underbrace{1\cdots1}_{k-2}04,

whose digit sum is

11+(k-2)

By carrying out the multiplication,

8\cdot\underbrace{88\ldots8}_{k}=7\underbrace{1\ldots1}_{k-2}04,

which has

k-2

ones.

The digit sum is

7+(k-2)+0+4=k+9.

Setting

k+9=1000

gives

k=991.

Thus, the correct answer is D.

Problem 16 in other years