# 2005 AMC 12B Problems/Problem 17

## Problem

How many distinct four-tuples $(a,b,c,d)$ of rational numbers are there with $$a\cdot\log_{10}2+b\cdot\log_{10}3+c\cdot\log_{10}5+d\cdot\log_{10}7=2005?$$ $\mathrm{(A)}\ 0 \qquad \mathrm{(B)}\ 1 \qquad \mathrm{(C)}\ 17 \qquad \mathrm{(D)}\ 2004 \qquad \mathrm{(E)}\ \text{infinitely many}$

## Solution

Using the laws of logarithms, the given equation becomes $$\log_{10}2^{a}+\log_{10}3^{b}+\log_{10}5^{c}+\log_{10}7^{d}=2005$$ $$\Rightarrow \log_{10}{2^{a}\cdot 3^{b}\cdot 5^{c}\cdot 7^{d}}=2005$$ $$\Rightarrow 2^{a}\cdot 3^{b}\cdot 5^{c}\cdot 7^{d} = 10^{2005}$$

As $a,b,c,d$ must all be rational, and there are no powers of $3$ or $7$ in $10^{2005}$, $b=d=0$. Then $2^{a}\cdot 5^{c}=2^{2005}\cdot 5^{2005} \Rightarrow a=c=2005$.

Only the four-tuple $(2005,0,2005,0)$ satisfies the equation, so the answer is $\boxed{1} \Rightarrow \mathrm{(B)}$.

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