1952 AMC 12 第 44 题

Question

若一个两位整数等于其各位数字之和的

k

倍，则将两位数字对调后所得的数等于各位数字之和乘以：

If an integer of two digits is

k

times the sum of its digits, the number formed by interchanging the digits is the sum of the digits multiplied by:

(9-k)

(10-k)

(11-k)

(k-1)

(k+1)

设两位数字为

a

和

b

，并表示原数与倒序数

Let the two digits be

a

and

b

, and express both the original and reversed numbers

两个数之和为

11(a+b)

The two numbers add to

11(a+b)

设原数为

10a+b=k(a+b)

。倒序数为

10b+a

。因为

\begin{gathered} (10a+b)+(10b+a) \\ =11(a+b) \end{gathered}\text{，}

所以倒序数等于

\begin{aligned} &11(a+b)-k(a+b)\\ &\qquad=(11-k)(a+b) \end{aligned}\text{。}

因此，正确答案是 C。

Let the original number be

10a+b=k(a+b).

The reversed number is

10b+a.

Since

\begin{gathered} (10a+b)+(10b+a) \\ =11(a+b), \end{gathered}

the reversed number equals

\begin{aligned} &11(a+b)-k(a+b)\\ &\qquad=(11-k)(a+b). \end{aligned}

Thus, the correct answer is C.

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