AMC12 2011 B

AMC12 2011 B · Q4

AMC12 2011 B · Q4. It mainly tests Primes & prime factorization, Digit properties (sum of digits, divisibility tests).

In multiplying two positive integers $a$ and $b$, Ron reversed the digits of the two-digit number $a$. His erroneous product was $161.$ What is the correct value of the product of $a$ and $b$?

在将两个正整数 $a$ 和 $b$ 相乘时，Ron 把两位数 $a$ 的数字顺序颠倒了。他得到的错误乘积是 $161$。$a$ 与 $b$ 的正确乘积是多少？

(A) 116 116

(B) 161 161

(D) 214 214

(E) 224 224

Answer

Correct choice: (E)

正确答案：（E）

Solution

Taking the prime factorization of $161$ reveals that it is equal to $23*7.$ Therefore, the only ways to represent $161$ as a product of two positive integers is $161*1$ and $23*7.$ Because neither $161$ nor $1$ is a two-digit number, we know that $a$ and $b$ are $23$ and $7.$ Because $23$ is a two-digit number, we know that a, with its two digits reversed, gives $23.$ Therefore, $a = 32$ and $b = 7.$ Multiplying our two correct values of $a$ and $b$ yields \[a*b = 32*7 =\] \[= \boxed{224 \textbf{(E)}}\]

对 $161$ 做质因数分解可得 $161=23*7$。因此，把 $161$ 表示为两个正整数的乘积只有 $161*1$ 和 $23*7$ 两种。由于 $161$ 和 $1$ 都不是两位数，可知 $a$ 和 $b$ 分别是 $23$ 和 $7$。因为 $23$ 是两位数，且将 $a$ 的两位数字颠倒后得到 $23$，所以 $a=32$，$b=7$。因此 \[a*b = 32*7 =\] \[= \boxed{224 \textbf{(E)}}\]

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