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 17:13, 15 February 2014
Problem
The fraction where
is the length of the period of the repeating decimal expansion. What is the sum
?
Solution
the fraction can be written as
.
similarly the fraction
can be written as
which is equivalent to
and we can see that for each
there are
combinations so the above sum is equivalent to:
we note that the sequence starts repeating at
yet consider
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
subtracting the sum of the digits of 98 which is 17 we get
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.