AMC12 2005 B

AMC12 2005 B · Q17

AMC12 2005 B · Q17. It mainly tests Logarithms (rare), Primes & prime factorization.

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

有多少个不同的有理数四元组$(a,b,c,d)$满足 \[a\cdot\log_{10}2+b\cdot\log_{10}3+c\cdot\log_{10}5+d\cdot\log_{10}7=2005?\]

(A) 0 0

(B) 1 1

(D) 2004 2004

(E) infinitely many 无限多个

Answer

Correct choice: (B)

正确答案：（B）

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 $\boxed{1} \Rightarrow \mathrm{(B)}$.

利用对数运算律，原方程化为 \[\log_{10}2^{a}+\log_{10}3^{b}+\log_{10}5^{c}+\log_{10}7^{d}=2005\] \[\Rightarrow \log_{10}{2^{a}\cdot 3^{b}\cdot 5^{c}\cdot 7^{d}}=2005\] \[\Rightarrow 2^{a}\cdot 3^{b}\cdot 5^{c}\cdot 7^{d} = 10^{2005}\] 由于$a,b,c,d$都必须是有理数，而$10^{2005}$中不含$3$或$7$的因子，所以$b=d=0$。于是$2^{a}\cdot 5^{c}=2^{2005}\cdot 5^{2005} \Rightarrow a=c=2005$。只有四元组$(2005,0,2005,0)$满足方程，因此答案为$\boxed{1} \Rightarrow \mathrm{(B)}$。

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