Difference between revisions of "2002 AIME I Problems/Problem 4"
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m (→Solution) |
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Which is | Which is | ||
− | <math>\dfrac{m | + | <math>\dfrac{n-m}{mn}=\dfrac{1}{29}</math> |
Since we need a 29 in the denominator, let <math>m=29t</math>.* Substituting, | Since we need a 29 in the denominator, let <math>m=29t</math>.* Substituting, | ||
− | <math>29t- | + | <math>29t-m=mt</math> |
<math>\frac{29t}{t+1} = n</math> | <math>\frac{29t}{t+1} = n</math> | ||
− | Since n is an integer, <math>t+1 = 29</math>, or <math>t=28</math>. It quickly follows that <math> | + | Since n is an integer, <math>t+1 = 29</math>, or <math>t=28</math>. It quickly follows that <math>n=29(28)</math> and <math>m=28</math>, so <math>m+n = 30(28) = \fbox{840}</math>. |
− | *If <math> | + | *If <math>m=29t</math>, a similar argument to the one above implies <math>m=29(28)</math> and <math>m=28</math>, which implies <math>m>n</math>, which is impossible since <math>n-m>0</math>. |
== See also == | == See also == | ||
{{AIME box|year=2002|n=I|num-b=3|num-a=5}} | {{AIME box|year=2002|n=I|num-b=3|num-a=5}} |
Revision as of 17:34, 29 June 2010
Problem
Consider the sequence defined by for
. Given that
, for positive integers
and
with
, find
.
Solution
. Thus,
Which is
Since we need a 29 in the denominator, let .* Substituting,
Since n is an integer, , or
. It quickly follows that
and
, so
.
- If
, a similar argument to the one above implies
and
, which implies
, which is impossible since
.
See also
2002 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |