Difference between revisions of "2016 AIME II Problems/Problem 9"
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Using the same reasoning (<math>100</math> isn't very big), we can guess which terms will work. The first case is <math>k=3</math>, so the second and fourth terms of <math>c</math> are <math>100</math> and <math>1000</math>. We let <math>r</math> be the common ratio of the geometric sequence and write the arithmetic terms in terms of <math>r</math>. | Using the same reasoning (<math>100</math> isn't very big), we can guess which terms will work. The first case is <math>k=3</math>, so the second and fourth terms of <math>c</math> are <math>100</math> and <math>1000</math>. We let <math>r</math> be the common ratio of the geometric sequence and write the arithmetic terms in terms of <math>r</math>. | ||
− | The common difference is <math>100-r - 1</math>, and so we can equate: <math>2(99-r)+100-r=1000-r^ | + | The common difference is <math>100-r - 1</math>, and so we can equate: <math>2(99-r)+100-r=1000-r^3</math>. Moving all the terms to one side and the constants to the other yields |
− | <math>r^ | + | <math>r^3-3r = 702</math>, so <math>r(r^2-3) = 702</math>. Simply listing out the factors of <math>702</math> shows that the only factor <math>3</math> less than a square that works is <math>78</math>. Thus <math>r=9</math> and we solve to get <math>\boxed{262}</math> from there. |
Solution by rocketscience | Solution by rocketscience |
Revision as of 22:56, 17 March 2016
The sequences of positive integers and
are an increasing arithmetic sequence and an increasing geometric sequence, respectively. Let
. There is an integer
such that
and
. Find
.
Solution
Since all the terms of the sequences are integers, and 100 isn't very big, we should just try out the possibilities for . When we get to
and
, we have
and
, which works, therefore, the answer is
.
Solution by Shaddoll
Solution 2
Using the same reasoning ( isn't very big), we can guess which terms will work. The first case is
, so the second and fourth terms of
are
and
. We let
be the common ratio of the geometric sequence and write the arithmetic terms in terms of
.
The common difference is , and so we can equate:
. Moving all the terms to one side and the constants to the other yields
, so
. Simply listing out the factors of
shows that the only factor
less than a square that works is
. Thus
and we solve to get
from there.
Solution by rocketscience