1952 AHSME Problems/Problem 44

Problem

If an integer of two digits is $k$ times the sum of its digits, the number formed by interchanging the 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 $n = 10a+b$. The problem states that $10a+b=k(a+b)$. We want to find $x$, where $10b+a=x(a+b)$. Adding these two equations gives $11(a+b) = (k+x)(a+b)$. Because $a+b \neq 0$, we have $11 = k + x$, or $x = \boxed{\textbf{(C) \ } 11-k}$.

See Also

1952 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 43
Followed by
Problem 45
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