Difference between revisions of "2014 AMC 12A Problems/Problem 23"
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similarly the fraction <math>\dfrac{1}{99^2}</math> can be written as <math>\sum^{\infty}_{m=1}\dfrac{1}{10^{2m}}\sum^{\infty}_{n=1}\dfrac{1}{10^{2n}}</math> which is equivalent to <cmath>\sum^{\infty}_{m=1}\sum^{\infty}_{n=1} \dfrac{1}{10^{2(m+n)}}</cmath> | similarly the fraction <math>\dfrac{1}{99^2}</math> can be written as <math>\sum^{\infty}_{m=1}\dfrac{1}{10^{2m}}\sum^{\infty}_{n=1}\dfrac{1}{10^{2n}}</math> which is equivalent to <cmath>\sum^{\infty}_{m=1}\sum^{\infty}_{n=1} \dfrac{1}{10^{2(m+n)}}</cmath> | ||
and we can see that for each <math>n+m=k</math> there are <math>k-1</math> <math>(n,m)</math> combinations so the above sum is equivalent to: | and we can see that for each <math>n+m=k</math> there are <math>k-1</math> <math>(n,m)</math> combinations so the above sum is equivalent to: | ||
| − | <cmath>\sum^{\infty}_{k=2}\dfrac{k-1}{10^{ | + | <cmath>\sum^{\infty}_{k=2}\dfrac{k-1}{10^{2k}}</cmath> |
we note that the sequence starts repeating at <math>k = 102</math> | we note that the sequence starts repeating at <math>k = 102</math> | ||
yet consider <cmath>\sum^{101}_{k=99}\dfrac{k-1}{10^{-2k}}=\dfrac{98}{{10^{198}}}+\dfrac{99}{{10^{200}}}+\dfrac{100}{10^{{202}}}=\dfrac{1}{10^{198}}(98+\dfrac{99}{100}+\dfrac{100}{10000})=\dfrac{1}{10^{198}}(98+\dfrac{99}{100}+\dfrac{1}{100})=\dfrac{1}{10^{198}}(98+\dfrac{100}{100})=\dfrac{1}{10^{198}}(99)</cmath> | yet consider <cmath>\sum^{101}_{k=99}\dfrac{k-1}{10^{-2k}}=\dfrac{98}{{10^{198}}}+\dfrac{99}{{10^{200}}}+\dfrac{100}{10^{{202}}}=\dfrac{1}{10^{198}}(98+\dfrac{99}{100}+\dfrac{100}{10000})=\dfrac{1}{10^{198}}(98+\dfrac{99}{100}+\dfrac{1}{100})=\dfrac{1}{10^{198}}(98+\dfrac{100}{100})=\dfrac{1}{10^{198}}(99)</cmath> | ||
Revision as of 16:13, 15 February 2014
Problem
The fraction ![\[\dfrac1{99^2}=0.\overline{b_{n-1}b_{n-2}\ldots b_2b_1b_0},\]](http://latex.artofproblemsolving.com/2/e/1/2e1249826c419acf9b5a00eb3f3f9acb191cc1d1.png)
where 
is the length of the period of the repeating decimal expansion. What is the sum 
?
Solution
the fraction 
can be written as ![\[\sum^{\infty}_{n=1}\dfrac{1}{10^{2n}}\]](http://latex.artofproblemsolving.com/4/3/9/4390881e118c5b4c8240b842b466517591116226.png)
.
similarly the fraction 
can be written as 
which is equivalent to ![\[\sum^{\infty}_{m=1}\sum^{\infty}_{n=1} \dfrac{1}{10^{2(m+n)}}\]](http://latex.artofproblemsolving.com/5/f/a/5fa6acb4ddde579cb4a0ed9a8338a9aad195b09d.png)
and we can see that for each 
there are 
combinations so the above sum is equivalent to:
![\[\sum^{\infty}_{k=2}\dfrac{k-1}{10^{2k}}\]](http://latex.artofproblemsolving.com/0/6/d/06db84aab337411b42a9811c3454520c711583f7.png)
we note that the sequence starts repeating at 
yet consider ![\[\sum^{101}_{k=99}\dfrac{k-1}{10^{-2k}}=\dfrac{98}{{10^{198}}}+\dfrac{99}{{10^{200}}}+\dfrac{100}{10^{{202}}}=\dfrac{1}{10^{198}}(98+\dfrac{99}{100}+\dfrac{100}{10000})=\dfrac{1}{10^{198}}(98+\dfrac{99}{100}+\dfrac{1}{100})=\dfrac{1}{10^{198}}(98+\dfrac{100}{100})=\dfrac{1}{10^{198}}(99)\]](http://latex.artofproblemsolving.com/e/5/9/e5975bc23a59e302b1666e966947d23a436dde56.png)
so the decimal will go from 1 to 99 skipping the number 98
and we can easily compute the sum of the digits from 0 to 99 to be ![\[45\cdot10\cdot2=900\]](http://latex.artofproblemsolving.com/a/b/8/ab8befb03ae1a73d638ebde336a7967cd10c8f82.png)
subtracting the sum of the digits of 98 which is 17 we get
![\[900-17=883\textbf{(B) }\qquad\]](http://latex.artofproblemsolving.com/1/f/8/1f865fdf8ef47666208820fda82567e614f0df96.png)
See Also
| 2014 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 22 |
Followed by Problem 24 |
| 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 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
