1952 AHSME Problems/Problem 43

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

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 $\fbox{C}$ .

1952 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 42	Followed by Problem 44
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1952 AHSME Problems/Problem 43

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