Difference between revisions of "2016 AMC 12B Problems/Problem 22"
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Solution by e_power_pi_times_i | Solution by e_power_pi_times_i | ||
+ | If <math>\frac{1}{n} = 0.\overline{abcdef}</math>, <math>n</math> must be a factor of <math>999999</math>. Also, by the same procedure, <math>n+6</math> must be a factor of <math>9999</math>. Checking through all the factors of <math>999999</math> and <math>9999</math> that are less than <math>1000</math>, we see that <math>n = 297</math> is a solution, so the answer is <math>\boxed{\textbf{(B)}}</math>. | ||
− | + | Note: <math>n = 27</math> is also a solution, which invalidates this method. However, we need to examine all factors of <math>999999</math> that are not factors of <math>99999</math>, <math>999</math>, or <math>99</math>, or <math>9</math>. Additionally, we need <math>n+6</math> to be a factor of <math>9999</math> but not <math>999</math>, <math>99</math>, or <math>9</math>. Indeed, <math>297</math> satisfies these requirements. | |
==See Also== | ==See Also== | ||
{{AMC12 box|year=2016|ab=B|num-b=21|num-a=23}} | {{AMC12 box|year=2016|ab=B|num-b=21|num-a=23}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 01:18, 23 February 2016
Problem
For a certain positive integer less than
, the decimal equivalent of
is
, a repeating decimal of period of
, and the decimal equivalent of
is
, a repeating decimal of period
. In which interval does
lie?
Solution
Solution by e_power_pi_times_i
If ,
must be a factor of
. Also, by the same procedure,
must be a factor of
. Checking through all the factors of
and
that are less than
, we see that
is a solution, so the answer is
.
Note: is also a solution, which invalidates this method. However, we need to examine all factors of
that are not factors of
,
, or
, or
. Additionally, we need
to be a factor of
but not
,
, or
. Indeed,
satisfies these requirements.
See Also
2016 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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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