Difference between revisions of "2009 AIME I Problems/Problem 7"
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<cmath>5^{a_n}=3n+2.</cmath> | <cmath>5^{a_n}=3n+2.</cmath> | ||
We can now test powers of <math>5</math>. | We can now test powers of <math>5</math>. | ||
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<math>5</math>-that gives us <math>n=1</math>, which is useless. | <math>5</math>-that gives us <math>n=1</math>, which is useless. | ||
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<math>25</math>-that gives a non-integer <math>n</math>. | <math>25</math>-that gives a non-integer <math>n</math>. | ||
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<math>125</math>-that given <math>n=\boxed{41}</math>. | <math>125</math>-that given <math>n=\boxed{41}</math>. | ||
Revision as of 20:51, 30 November 2022
Contents
[hide]Problem
The sequence satisfies
and
for
. Let
be the least integer greater than
for which
is an integer. Find
.
Solution
The best way to solve this problem is to get the iterated part out of the exponent:
Plug in
to see the first few terms of the sequence:
We notice that the terms
are in arithmetic progression. Since
, we can easily use induction to show that
. So now we only need to find the next value of
that makes
an integer. This means that
must be a power of
. We test
:
This has no integral solutions, so we try
:
Solution 2 (Telescoping?)
We notice that by multiplying the equation from an arbitrary all the way to
, we get:
This simplifies to
We can now test powers of
.
-that gives us
, which is useless.
-that gives a non-integer
.
-that given
.
-integralarefun
See also
2009 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.