AMC12 2006 B

AMC12 2006 B · Q22

AMC12 2006 B · Q22. It mainly tests Primes & prime factorization, Inequalities with integers (floor/ceiling basics).

Suppose $a$, $b$ and $c$ are positive integers with $a+b+c=2006$, and $a!b!c!=m\cdot 10^n$, where $m$ and $n$ are integers and $m$ is not divisible by $10$. What is the smallest possible value of $n$?

设 $a$、$b$ 和 $c$ 是正整数，且 $a+b+c=2006$，并且 $a!b!c!=m\cdot 10^n$，其中 $m$ 和 $n$ 为整数且 $m$ 不被 $10$ 整除。$n$ 的最小可能值是多少？

(A) 489 489

(B) 492 492

(D) 498 498

(E) 501 501

Answer

Correct choice: (B)

正确答案：（B）

Solution

The power of $10$ for any factorial is given by the well-known algorithm \[\left\lfloor \frac n{5}\right\rfloor + \left\lfloor \frac n{25}\right\rfloor + \left\lfloor \frac n{125}\right\rfloor + \cdots\] It is rational to guess numbers right before powers of $5$ because we won't have any extra numbers from higher powers of $5$. As we list out the powers of 5, it is clear that $5^{4}=625$ is less than 2006 and $5^{5}=3125$ is greater. Therefore, set $a$ and $b$ to be 624. Thus, c is $2006-(624\cdot 2)=758$. Applying the algorithm, we see that our answer is $152+152+188= \boxed{492}$.

任意阶乘中 $10$ 的幂次由著名算法给出： \[\left\lfloor \frac n{5}\right\rfloor + \left\lfloor \frac n{25}\right\rfloor + \left\lfloor \frac n{125}\right\rfloor + \cdots\] 合理的猜测是取接近 $5$ 的幂次的前一个数，因为这样不会从更高次的 $5$ 的幂中额外得到数。列出 $5$ 的幂可知 $5^{4}=625$ 小于 $2006$，而 $5^{5}=3125$ 大于 $2006$。因此令 $a$ 和 $b$ 都为 $624$，则 $c=2006-(624\cdot 2)=758$。应用上述算法可得答案为 $152+152+188= \boxed{492}$。

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