AMC10 2009 A

AMC10 2009 A · Q25

AMC10 2009 A · Q25. It mainly tests Primes & prime factorization, Counting divisors.

For $k > 0$, let $I_k = 10\dots064$, 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 $N(k)$?

对于$k>0$，令$I_k=10\dots064$，其中1和6之间有$k$个零。$N(k)$是$I_k$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

Answer (B): Note that $I_k = 2^{k+2}\cdot 5^{k+2} + 2^6$. For $k<4$, the first term is not divisible by $2^6$, so $N(k)<6$. For $k>4$, the first term is divisible by $2^7$, but the second term is not, so $N(k)<7$. For $k=4$, $I_4 = 2^6(5^6+1)$, and because the second factor is even, $N(4)\ge 7$. In fact the second factor is a sum of cubes so $$(5^6+1)=((5^2)^3+1^3)=(5^2+1)((5^2)^2-5^2+1).$$ The factor $5^2+1=26$ is divisible by $2$ but not $4$, and the second factor is odd, so $5^6+1$ contributes one more factor of $2$. Hence the maximum value for $N(k)$ is $7$.

答案（B）：注意 $I_k = 2^{k+2}\cdot 5^{k+2} + 2^6$。当 $k<4$ 时，第一项不被 $2^6$ 整除，所以 $N(k)<6$。当 $k>4$ 时，第一项可被 $2^7$ 整除，但第二项不行，所以 $N(k)<7$。当 $k=4$ 时，$I_4 = 2^6(5^6+1)$，且因为第二个因子是偶数，所以 $N(4)\ge 7$。实际上第二个因子是立方和，因此 $$(5^6+1)=((5^2)^3+1^3)=(5^2+1)((5^2)^2-5^2+1).$$ 因子 $5^2+1=26$ 可被 $2$ 整除但不可被 $4$ 整除，而第二个因子是奇数，所以 $5^6+1$ 只再贡献一个 $2$ 的因子。因此 $N(k)$ 的最大值为 $7$。

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