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

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== Solution ==
 
== Solution ==
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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>.
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<math>\boxed{}</math>
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Setting AB equal to <math>k</math> times the sum of the digits yields
 
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<cmath>10A + B = k(A + B)</cmath>
 
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We now must relate AB to BA. Note that
 
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<cmath>11(A + B) - (10A + B) = 10B + A</cmath>
 
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Using this in the first equation yields
 
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<cmath>10A + B = k(A + B)</cmath>
 
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<cmath>11(A + B) - (10A + B) = 11(A + B) - k(A + B)</cmath>
 
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<cmath>10B + A = (11 - k)(A + B)</cmath>
 
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Therefore, <math>BA</math> is <math>(11 - k)</math> times the sum of its digits and our answer is <math>\fbox{C}</math>.
 
  
 
== See also ==
 
== See also ==

Revision as of 07:41, 1 May 2016

Problem

The diameter of a circle is divided into $n$ equal parts. On each part a semicircle is constructed. As $n$ becomes very large, the sum of the lengths of the arcs of the semicircles approaches a length: $\textbf{(A) } \qquad$ equal to the semi-circumference of the original circle $\textbf{(B) } \qquad$ equal to the diameter of the original circle $\textbf{(C) } \qquad$ greater than the diameter, but less than the semi-circumference of the original circle $\textbf{(D) } \qquad$ that is infinite $\textbf{(E) }$ greater than the semi-circumference

Solution

$\boxed{}$

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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