2014 AMC 12A Problem 16

Below is the professionally curated solution for Problem 16 of the 2014 AMC 12A, 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.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:digitspattern recognitionsmall cases

Difficulty rating: 1660

16.

The product (8)(8888),(8)(888\ldots8), where the second factor has kk digits, is an integer whose digits have a sum of 1000.1000. What is k?k?

901901

911911

919919

991991

999999

Solution:

By carrying out the multiplication, 8888k=711k204,8\cdot\underbrace{88\ldots8}_{k}=7\underbrace{1\ldots1}_{k-2}04, which has k2k-2 ones.

The digit sum is 7+(k2)+0+4=k+9.7+(k-2)+0+4=k+9. Setting k+9=1000k+9=1000 gives k=991.k=991.

Thus, the correct answer is D.

Problem 16 in Other Years