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2009 AMC 12A Problem 12

Problem 12 of 25IntermediateNumber TheoryProblem-Solving Techniques

How many positive integers less than 10001000 are 66 times the sum of their digits?

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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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