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2005 AMC 12B Problem 17

Problem 17 of 25IntermediateAlgebraNumber Theory

How many distinct four-tuples (a,b,c,d)(a, b, c, d) of rational numbers are there with alog102+blog103+clog105+dlog107=2005? \begin{aligned} &a\log_{10} 2 + b\log_{10} 3 \\ &\quad {}+ c\log_{10} 5 + d\log_{10} 7 = 2005? \end{aligned}

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Solution

The equation is equivalent to log10(2a3b5c7d)=2005,\log_{10}\left(2^a 3^b 5^c 7^d\right) = 2005, so 2a3b5c7d=102005=2200552005. 2^a 3^b 5^c 7^d = 10^{2005} = 2^{2005} \cdot 5^{2005}. Clearing the denominators of a,b,c,da, b, c, d with a common integer multiplier and using the uniqueness of prime factorization, the exponents must match: a=2005,a = 2005, b=0,b = 0, c=2005,c = 2005, and d=0.d = 0. So there is exactly 11 such four-tuple. Thus, the correct answer is B.

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Concepts: logarithm · prime factorization

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