Difference between revisions of "1952 AHSME Problems/Problem 43"
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<math>\textbf{(A) } \qquad</math> equal to the semi-circumference of the original circle | <math>\textbf{(A) } \qquad</math> equal to the semi-circumference of the original circle | ||
<math>\textbf{(B) } \qquad</math> equal to the diameter of the original circle | <math>\textbf{(B) } \qquad</math> equal to the diameter of the original circle | ||
| − | <math>\textbf{(C) } | + | <math>\textbf{(C) } \qquad</math> greater than the diameter, but less than the semi-circumference of the original circle |
| − | <math>\textbf{(D) } | + | <math>\textbf{(D) } \qquad</math> that is infinite |
| − | <math>\textbf{(E) } | + | <math>\textbf{(E) } </math> greater than the semi-circumference |
== Solution == | == Solution == | ||
Revision as of 07:40, 1 May 2016
Problem
The diameter of a circle is divided into 
equal parts. On each part a semicircle is constructed. As becomes very large, the sum of the lengths of the arcs of the semicircles approaches a length:
equal to the semi-circumference of the original circle
equal to the diameter of the original circle
greater than the diameter, but less than the semi-circumference of the original circle
that is infinite
greater than the semi-circumference
Solution
Let our two digit number be 
. Its value is 
. The number formed by interchanging its digits is BA and has value 
.
Setting AB equal to 
times the sum of the digits yields
![\[10A + B = k(A + B)\]](http://latex.artofproblemsolving.com/3/5/0/35087c9b4da7b9cc6c723ed0798ac90c3ed8f4fc.png)
We now must relate AB to BA. Note that
![\[11(A + B) - (10A + B) = 10B + A\]](http://latex.artofproblemsolving.com/7/b/6/7b6ac880a0b6ecdd549728e2b935ebe738a596e6.png)
Using this in the first equation yields
![\[11(A + B) - (10A + B) = 11(A + B) - k(A + B)\]](http://latex.artofproblemsolving.com/8/e/3/8e3f54021fc96aa88ee57a2a7f16e87ae0f35dd7.png)
![\[10B + A = (11 - k)(A + B)\]](http://latex.artofproblemsolving.com/c/1/9/c19d19e483c444c3f32922a1def767daa2fa896f.png)
Therefore, 
is 
times the sum of its digits and our answer is 
.
See also
| 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. 
