Difference between revisions of "2014 AMC 12A Problems/Problem 23"
(→Solution 2) |
(→Solution 2) |
||
Line 26: | Line 26: | ||
=\frac{0.\overline{01}}{99}\\\\ | =\frac{0.\overline{01}}{99}\\\\ | ||
=0.\overline{00010203...9799}</math> | =0.\overline{00010203...9799}</math> | ||
+ | \\\\So the answer is <math>00+01+02+03+...+97+99=/frac{99\cdot100}{2}-98</math> or <math>(B) 883</math>. | ||
==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 18:58, 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
\\\\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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.