Difference between revisions of "2016 AMC 12B Problems/Problem 22"
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Note: <math>n = 27</math> and <math>n = 3</math> are both solutions, 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. | Note: <math>n = 27</math> and <math>n = 3</math> are both solutions, 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. | ||
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+ | We can see that <math>n=27</math> and <math>n=3</math> are not solutions by checking it in the requirements of the problem: <math>\frac{1}{3}=0.3333\dots</math>, period 1, and <math>\frac{1}{27}=0.037037\dots</math>, period 3. Thus, <math>n=297</math> is the only answer. | ||
For anyone who wants more information about repeating decimals, visit: https://en.wikipedia.org/wiki/Repeating_decimal | For anyone who wants more information about repeating decimals, visit: https://en.wikipedia.org/wiki/Repeating_decimal |
Revision as of 11:57, 17 April 2019
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: and
are both solutions, 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.
We can see that and
are not solutions by checking it in the requirements of the problem:
, period 1, and
, period 3. Thus,
is the only answer.
For anyone who wants more information about repeating decimals, visit: https://en.wikipedia.org/wiki/Repeating_decimal
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.