Difference between revisions of "2014 AMC 12A Problems/Problem 23"
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==Solution 1== | ==Solution 1== | ||
− | + | The fraction <math>\dfrac{1}{99}</math> can be written as <cmath>\sum^{\infty}_{n=1}\dfrac{1}{10^{2n}}</cmath>. | |
− | + | Similarly, thy 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^{2k}}</cmath> | <cmath>\sum^{\infty}_{k=2}\dfrac{k-1}{10^{2k}}</cmath> |
Revision as of 15:13, 16 January 2023
Problem
The fraction
where is the length of the period of the repeating decimal expansion. What is the sum
?
Solution 1
The fraction can be written as
.
Similarly, thy 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
Solution 2
So, the answer is or
.
There are two things to notice here. First, has a very simple and unique decimal expansion, as shown. Second, for
to itself produce a repeating decimal,
has to evenly divide a sufficiently extended number of the form
. This number will have
ones (197 digits in total), as to be divisible by
and
. The enormity of this number forces us to look for a pattern, and so we divide out as shown. Indeed, upon division (seeing how the remainders always end in "501" or "601" or, at last, "9801"), we find the repeating part
. If we wanted to further check our pattern, we could count the total number of digits in our quotient (not counting the first three): 195. Since
, multiplying by it will produce either
or
extra digits, so our quotient passes the test.
Or note that the 1 has to be carried when you get to 100 so the 99 becomes 00 and the 1 gets carried again to 98 which becomes 99.
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2014amc12a/382
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.