2009 AMC 12A Problem 12

Attempt Problem 12 of the 2009 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 2009 AMC 12A solutions, or check the answer key.

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

12.

How many positive integers less than 10001000 are 66 times the sum of their digits?
00
11
22
44
1212
Answer: B
Concepts:digitsbounding to limit casescasework
Small hint:
If NN equals 66 times its digit sum, then NN is a multiple of 66 and NN is small
Big hint:
The digit sum of a number below 10001000 is at most 27,27, so N≤162N \le 162
Solution:
If N=6⋅(digit sum),N = 6\cdot(\text{digit sum}), then since the digit sum of a number below 10001000 is at most 27,27, we have N≤162.N \le 162.
For a two-digit number 10t+u=6(t+u)10t + u = 6(t + u) gives 4t=5u,4t = 5u, forcing t=5t = 5 and u=4,u = 4, so N=54.N = 54. A one-digit number would need 6u=u,6u = u, impossible for u>0.u \gt 0. A three-digit number 100h+10t+u=6(h+t+u)100h + 10t + u = 6(h + t + u) gives 94h+4t=5u,94h + 4t = 5u, whose left side is at least 9494 while the right side is at most 45,45, so there is no solution.
Hence exactly one number, 54,54, works.
Thus, the correct answer is B.
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