AMC10 2014 B

AMC10 2014 B · Q17

AMC10 2014 B · Q17. It mainly tests Primes & prime factorization, Parity (odd/even).

What is the greatest power of 2 that is a factor of $10^{1002} - 4^{501}$?

$10^{1002}-4^{501}$的最大2的幂因数是多少？

(A) $2^{1002}$ $2^{1002}$

(B) $2^{1003}$ $2^{1003}$

(D) $2^{1005}$ $2^{1005}$

(E) $2^{1006}$ $2^{1006}$

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Note that $10^{1002}-4^{501}=2^{1002}\cdot 5^{1002}-2^{1002}$ $=2^{1002}(5^{1002}-1)$ $=2^{1002}(5^{501}-1)(5^{501}+1)$ $=2^{1002}(5-1)(5^{500}+5^{499}+\cdots+5+1)(5+1)(5^{500}-5^{499}+\cdots-5+1)$ $=2^{1005}(3)(5^{500}+5^{499}+\cdots+5+1)(5^{500}-5^{499}+\cdots-5+1).$ Because each of the last two factors is a sum of an odd number of odd terms, they are both odd. The greatest power of 2 is $2^{1005}$.

答案 (D)：注意 $10^{1002}-4^{501}=2^{1002}\cdot 5^{1002}-2^{1002}$ $=2^{1002}(5^{1002}-1)$ $=2^{1002}(5^{501}-1)(5^{501}+1)$ $=2^{1002}(5-1)(5^{500}+5^{499}+\cdots+5+1)(5+1)(5^{500}-5^{499}+\cdots-5+1)$ $=2^{1005}(3)(5^{500}+5^{499}+\cdots+5+1)(5^{500}-5^{499}+\cdots-5+1).$ 因为最后两个因子分别是由奇数个奇数项相加得到的，所以它们都是奇数。2 的最大幂为 $2^{1005}$。

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