Difference between revisions of "1952 AHSME Problems/Problem 44"

Revision as of 11:31, 5 July 2013

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

1952 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 43	Followed by Problem 44
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All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 == See Also ==
 {{AHSME 50p box|year=1952|num-b=43|num-a=44}}
+{{MAA Notice}}

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Difference between revisions of "1952 AHSME Problems/Problem 44"

Revision as of 11:31, 5 July 2013

Problem

Solution

See Also