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2009 AMC 10B Problem 6

Problem 6 of 25EasierAlgebraNumber Theory

Kiana has two older twin brothers. The product of their three ages is 128.128. What is the sum of their three ages?

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Solution

Since 128=27,128=2^7, every age is a power of 2.2. Writing the twins’ common age as tt and Kiana’s as k,k, we need t2k=128t^2k=128 with k<t.k\lt t. Write t=2j.t=2^j. Then k=27−2j.k=2^{7-2j}. For kk to be a positive integer age we need 7−2j≥0,7-2j\ge0, so j≤3.j\le3. Because Kiana is younger than the twins, 7−2j<j,7-2j\lt j, so j≥3.j\ge3. Therefore j=3,j=3, giving t=8t=8 and k=2.k=2. The sum is 8+8+2=18.8+8+2=18. Thus, the correct answer is D.
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Tagged: prime factorization · power of 2 · ages

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