1952 AMC 12 Problem 44

Question

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)

Let the two digits be

a

and

b

, and express both the original and reversed numbers

The two numbers add to

11(a+b)

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.

Problem 44 in other years