1952 AHSME Problems/Problem 43

Problem

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:

$\textbf{(A) } (9 - k) \qquad \textbf{(B) } (10 - k) \qquad \textbf{(C) } (11 - k) \qquad \textbf{(D) } (k - 1) \qquad \textbf{(E) } (k+1)$

Solution

Let our two digit number be $AB$ . Its value in base 10 is $10A + B$ . The number formed by interchanging its digits is BA and has value $10B + A$ . Setting AB equal to $k$ times the sum of the digits yields $10A + B = k(A + B)$ We now must relate AB. Note that $11(A + B) - (10A + B) = 10B + A$ Using this in the first equation yields $10A + B = k(A + B)$ $11(A + B) - (10A + B) = 11(A + B) - k(A + B)$ $10B + A = (11 - k)(A + B)$ Therefore, $BA$ is $(11 - k)$ times the sum of its digits and our answer is $\fbox{C}$ .

1952 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 42	Followed by Problem 44
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1952 AHSME Problems/Problem 43

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