2014 AMC 12A Problems/Problem 23
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.
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 |
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