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:  
  
<math>\textbf{(A) } (9 - k) \qquad
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<math>\textbf{(A) } \qquad</math> equal to the semi-circumference of the original circle
\textbf{(B) } (10 - k) \qquad
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<math>\textbf{(B) } \qquad</math> equal to the diameter of the original circle
\textbf{(C) } (11 - k) \qquad
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<math>\textbf{(C) } \qquad</math> greater than the diameter, but less than the semi-circumference of the original circle
\textbf{(D) } (k - 1) \qquad
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<math>\textbf{(D) } \qquad</math> that is infinite
\textbf{(E) } (k+1) </math>
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<math>\textbf{(E) } </math> greater than the semi-circumference
  
 
== 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>.
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Note that the half the circumference of a circle with diameter <math>d</math> is <math>\frac{\pi*d}{2}</math>.  
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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Let's call the diameter of the circle D. Dividing the circle's diameter into n parts means that each semicircle has diameter <math>\frac{D}{n}</math>, and thus each semicircle measures <math>\frac{D*pi}{n*2}</math>. The total sum of those is <math>n*\frac{D*pi}{n*2}=\frac{D*pi}{2}</math>, and since that is the exact expression for the semi-circumference of the original circle, the answer is <math>\boxed{A}</math>.
We now must relate AB. Note that
 
<cmath>11(A + B) - (10A + B) = 10B + A</cmath>
 
Using this in the first equation yields
 
<cmath>10A + B = k(A + B)</cmath>
 
<cmath>11(A + B) - (10A + B) = 11(A + B) - k(A + B)</cmath>
 
<cmath>10B + A = (11 - k)(A + B)</cmath>
 
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 ==

Latest revision as of 11:28, 29 December 2024

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

Note that the half the circumference of a circle with diameter $d$ is $\frac{\pi*d}{2}$.

Let's call the diameter of the circle D. Dividing the circle's diameter into n parts means that each semicircle has diameter $\frac{D}{n}$, and thus each semicircle measures $\frac{D*pi}{n*2}$. The total sum of those is $n*\frac{D*pi}{n*2}=\frac{D*pi}{2}$, and since that is the exact expression for the semi-circumference of the original circle, the answer is $\boxed{A}$.

See also

1952 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 42
Followed by
Problem 44
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