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:

$(9-k)$

$(10-k)$

$(11-k)$

$(k-1)$

$(k+1)$

Answer: C

Concepts:digits algebraic manipulation

Difficulty rating: 1420

Small Hint:

Let the two digits be $a$ and $b$ , and express both the original and reversed numbers

Big Hint:

The two numbers add to $11(a+b)$

Solution:

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.

1952 AMC 12 Problem 44

Answer: C

Difficulty rating: 1420

Small Hint:

Big Hint:

Solution:

Problem 44 in Other Years