Difference between revisions of "2014 AMC 12A Problems/Problem 23"
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<math>\frac{1}{99^2}\\\\ | <math>\frac{1}{99^2}\\\\ | ||
=\frac{1}{99} \cdot \frac{1}{99}\\\\ | =\frac{1}{99} \cdot \frac{1}{99}\\\\ | ||
− | =\frac{0.\overline{01}}{99}</math> | + | =\frac{0.\overline{01}}{99}\\\\ |
+ | =0.\overline{00010203...9799}\\& | ||
+ | So the answer is the </math>00+01+02+03+ | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2014|ab=A|num-b=22|num-a=24}} | {{AMC12 box|year=2014|ab=A|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 19:52, 15 February 2014
Contents
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 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
Solution 2
$\frac{1}{99^2}\\\\ =\frac{1}{99} \cdot \frac{1}{99}\\\\ =\frac{0.\overline{01}}{99}\\\\ =0.\overline{00010203...9799}\\& So the answer is the$ (Error compiling LaTeX. Unknown error_msg)00+01+02+03+
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 |
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