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

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== Problem ==
 
== Problem ==
  
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If an integer of two digits is <math>k</math> times the sum of its digits, the number formed by interchanging the the digits is the sum of the digits multiplied by
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If an integer of two digits is <math>k</math> times the sum of its digits, the number formed by interchanging the digits is the sum of the digits multiplied by
  
 
<math> \textbf{(A) \ } 9-k  \qquad \textbf{(B) \ } 10-k \qquad \textbf{(C) \ } 11-k \qquad \textbf{(D) \ } k-1 \qquad \textbf{(E) \ } k+1 </math>
 
<math> \textbf{(A) \ } 9-k  \qquad \textbf{(B) \ } 10-k \qquad \textbf{(C) \ } 11-k \qquad \textbf{(D) \ } k-1 \qquad \textbf{(E) \ } k+1 </math>

Latest revision as of 03:38, 15 January 2023

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

See Also

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
All AHSME Problems and Solutions

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