Difference between revisions of "2014 AMC 12A Problems/Problem 23"
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− | the fraction < | + | the fraction <math>\dfrac{1}{99}</math> can be written as <cmath>\sum^{\infty}_{n=1}\dfrac{1}{10^{-2n}}</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> | 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: |
Revision as of 21:19, 7 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 is repeating for 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