2005 AMC 12B Problem 17

Below is the professionally curated solution for Problem 17 of the 2005 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2005 AMC 12B solutions, or check the answer key.

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Concepts:logarithm prime factorization

Difficulty rating: 1800

17.

How many distinct four-tuples $(a, b, c, d)$ of rational numbers are there with $a\log_{10} 2 + b\log_{10} 3 + c\log_{10} 5 + d\log_{10} 7 = 2005?$

$0$

$1$

$17$

$2004$

infinitely many

Solution:

The equation is equivalent to $\log_{10}\left(2^a 3^b 5^c 7^d\right) = 2005,$ so $2^a 3^b 5^c 7^d = 10^{2005} = 2^{2005} \cdot 5^{2005}.$

Clearing the denominators of $a, b, c, d$ with a common integer multiplier and using the uniqueness of prime factorization, the exponents must match: $a = 2005,$ $b = 0,$ $c = 2005,$ and $d = 0.$

So there is exactly $1$ such four-tuple.

Thus, the correct answer is B.

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