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 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 $x = \boxed{\textbf{(C) \ } 11-k}$ .

1952 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 43	Followed by Problem 45
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
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1952 AHSME Problems/Problem 44

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