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

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== Problem ==
 
== Problem ==
  
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:  
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The diameter of a circle is divided into <math>n</math> equal parts. On each part a semicircle is constructed. As <math>n</math> becomes very large, the sum of the lengths of the arcs of the semicircles approaches a length:  
 
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<math>\textbf{(A) } \qquad</math> equal to the semi-circumference of the original circle
<math>\textbf{(A) } (9 - k) \qquad
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<math>\textbf{(B) } \qquad</math> equal to the diameter of the original circle
\textbf{(B) } (10 - k) \qquad
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<math>\textbf{(C) } (11 - k) \qquad</math> greater than the diameter, but less than the semi-circumference of the original circle
\textbf{(C) } (11 - k) \qquad
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<math>\textbf{(D) } (k - 1) \qquad</math> that is infinite
\textbf{(D) } (k - 1) \qquad
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<math>\textbf{(E) } (k+1) </math> greater than the semi-circumference
\textbf{(E) } (k+1) </math>
 
  
 
== Solution ==
 
== Solution ==

Revision as of 08:40, 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 $\textbf{(C) } (11 - k) \qquad$ greater than the diameter, but less than the semi-circumference of the original circle $\textbf{(D) } (k - 1) \qquad$ that is infinite $\textbf{(E) } (k+1)$ greater than the semi-circumference

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 42
Followed by
Problem 44
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