AMC12 2009 A

AMC12 2009 A · Q18

AMC12 2009 A · Q18. It mainly tests Primes & prime factorization, Perfect squares & cubes.

For $k > 0$, let $I_k = 10\ldots 064$, where there are $k$ zeros between the $1$ and the $6$. Let $N(k)$ be the number of factors of $2$ in the prime factorization of $I_k$. What is the maximum value of $N(k)$?

对 $k>0$，令 $I_k=10\ldots064$，其中在 $1$ 与 $6$ 之间有 $k$ 个零。令 $N(k)$ 为 $I_k$ 的质因数分解中因子 $2$ 的个数。$N(k)$ 的最大值是多少？

(A) 6 6

(B) 7 7

(D) 9 9

(E) 10 10

Answer

Correct choice: (B)

正确答案：（B）

Solution

The number $I_k$ can be written as $10^{k+2} + 64 = 5^{k+2}\cdot 2^{k+2} + 2^6$. For $k\in\{1,2,3\}$ we have $I_k = 2^{k+2} \left( 5^{k+2} + 2^{4-k} \right)$. The first value in the parentheses is odd, the second one is even, hence their sum is odd and we have $N(k)=k+2\leq 5$. For $k>4$ we have $I_k=2^6 \left( 5^{k+2}\cdot 2^{k-4} + 1 \right)$. For $k>4$ the value in the parentheses is odd, hence $N(k)=6$. This leaves the case $k=4$. We have $I_4 = 2^6 \left( 5^6 + 1 \right)$. The value $5^6 + 1$ is obviously even. And as $5\equiv 1 \pmod 4$, we have $5^6 \equiv 1 \pmod 4$, and therefore $5^6 + 1 \equiv 2 \pmod 4$. Hence the largest power of $2$ that divides $5^6+1$ is $2^1$, and this gives us the desired maximum of the function $N$: $N(4) = \boxed{7}$.

数 $I_k$ 可写为 $10^{k+2}+64=5^{k+2}\cdot2^{k+2}+2^6$。当 $k\in\{1,2,3\}$ 时，$I_k=2^{k+2}\left(5^{k+2}+2^{4-k}\right)$。括号内第一项为奇数，第二项为偶数，因此和为奇数，故 $N(k)=k+2\le 5$。当 $k>4$ 时，$I_k=2^6\left(5^{k+2}\cdot2^{k-4}+1\right)$。此时括号内为奇数，故 $N(k)=6$。剩下 $k=4$ 的情况。此时 $I_4=2^6\left(5^6+1\right)$。$5^6+1$ 显然为偶数。又因为 $5\equiv1\pmod4$，所以 $5^6\equiv1\pmod4$，从而 $5^6+1\equiv2\pmod4$。因此能整除 $5^6+1$ 的最大 $2$ 的幂为 $2^1$，于是函数 $N$ 的最大值为 $N(4)=\boxed{7}$。

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