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 $\textbf{(C) } (11 - k) \qquad$ 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 $\fbox{C}$ .

1952 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 42	Followed by Problem 44
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1952 AHSME Problems/Problem 43

Problem

Solution

See also