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Difference between revisions of "1952 AHSME Problems/Problem 43"

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== Solution ==
 
== Solution ==
−
Let our two digit number be <math>AB</math>. Its value in base 10 is <math>10A + B</math>. The number formed by interchanging its digits is BA and has value <math>10B + A</math>.
+
Let our two digit number be <math>AB</math>. Its value is <math>10A + B</math>. The number formed by interchanging its digits is BA and has value <math>10B + A</math>.
 
Setting AB equal to <math>k</math> times the sum of the digits yields
 
Setting AB equal to <math>k</math> times the sum of the digits yields
 
<cmath>10A + B = k(A + B)</cmath>
 
<cmath>10A + B = k(A + B)</cmath>
−
We now must relate AB. Note that
+
We now must relate AB to BA. Note that
 
<cmath>11(A + B) - (10A + B) = 10B + A</cmath>
 
<cmath>11(A + B) - (10A + B) = 10B + A</cmath>
 
Using this in the first equation yields
 
Using this in the first equation yields

Revision as of 19:58, 22 December 2015

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 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 to BA. 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 $\fbox{C}$.

See also

1952 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 42 		Followed by
Problem 44
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All AHSME Problems and Solutions

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