2014 AMC 10A Problem 20

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Concepts:digitspattern recognitioninduction

Difficulty rating: 1660

20.

The product (8)(8888),(8)(888\dots8), 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:

To see if any pattern exists, we can test out small values of k.k.

We have that 88=64, 8 \cdot 8 = 64, 888=704, 8 \cdot 88 = 704, 8888=7104, 8 \cdot 888 = 7104, 88888=71104. 8 \cdot 8888 = 71104.

From this, it is pretty safe to guess that for every increment of k,k, there is an extra 11 added to the product. If you know how to use induction, you can prove that this pattern holds, but that's not necessary to solve the problem.

This means that for any k3,k \geq 3, the sum of the digits in the product is 7+4+0+k2=k+9. 7 + 4 + 0 + k - 2 = k + 9.

Finally, we get k+9=1000 k + 9 = 1000 k=991. k = 991.

Thus, D is the correct answer.

Problem 20 in Other Years