AMC8 2011

AMC8 2011 · Q17

AMC8 2011 · Q17. It mainly tests Primes & prime factorization.

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

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

(A) 21 21

(B) 25 25

(D) 35 35

(E) 56 56

Answer

Correct choice: (A)

正确答案：（A）

Solution

The prime factorization of $588$ is $2^2\cdot3\cdot7^2.$ We can see $w=2, x=1,$ and $z=2.$ Because $5^0=1, y=0.$ \[2w+3x+5y+7z=4+3+0+14=\boxed{\textbf{(A)}\ 21}\]

588的质因数分解为$2^2\cdot3\cdot7^2$。我们看到$w=2, x=1,$ 和 $z=2$。因为$5^0=1, y=0$。 \[2w+3x+5y+7z=4+3+0+14=\boxed{\textbf{(A)}\ 21}\]

Topics

NT.002 Primes & prime factorization