Difference between revisions of "2014 AMC 12A Problems/Problem 23"
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<math>=\frac{1}{99} \cdot \frac{1}{99}</math> | <math>=\frac{1}{99} \cdot \frac{1}{99}</math> | ||
<math>=\frac{0.\overline{01}}{99}</math> | <math>=\frac{0.\overline{01}}{99}</math> | ||
− | <math>=0.\overline{01}+0. | + | <math>=0.\overline{01}+0.00\overline{01}...</math> |
<math>=0.\overline{00010203...9799}</math> | <math>=0.\overline{00010203...9799}</math> | ||
Revision as of 13:07, 3 September 2017
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
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.
Solution 3 (Fastest)
Simpler version of the above solution
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.
So, the answer is or
.
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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