Question

设 $w$ 、 $x$ 、 $y$ 、 $z$ 为整数。若那么等于多少？ $2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588,$ $2w + 3x + 5y + 7z$

Let $w,$ $x,$ $y,$ and $z$ be whole numbers. If $2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588,$ then what does $2w + 3x + 5y + 7z$ equal?

$21$

$25$

$27$

$35$

$56$

为了找到所需指数，注意所有底数都是质数。因此质因数分解会有帮助。

有 $588 = 2^2 \cdot 3^1 \cdot 7^2.$

由此可知 $w = 2$ 、 $x = 1$ 、 $y = 0$ 且 $z = 2$ ，其中 $y = 0$ 使 $5^y$ 这一项等于 $1$ 。

代入这些指数可得 $\begin{gather*} 2w + 3x + 5y + 7z \\ = 2 \cdot 2 + 3 \cdot 1 + 5 \cdot 0 + 7 \cdot 2 \\ = 21. \end{gather*}$

所以正确答案是 A。

To find the desired exponents, note that all the bases are prime numbers. This means that finding the prime factorization will be helpful.

We get that $588 = 2^2 \cdot 3^1 \cdot 7^2.$

From this, it is clear that $w = 2,$ $x = 1,$ $y = 0,$ and $z = 2$ ( $y = 0$ since that makes the $5^y$ term equal $1$ ).

Therefore, $\begin{gather*} 2w + 3x + 5y + 7z \\ = 2 \cdot 2 + 3 \cdot 1 + 5 \cdot 0 + 7 \cdot 2 \\ = 21. \end{gather*}$

Thus, A is the correct answer.