Difference between revisions of "2002 AIME I Problems/Problem 4"
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<math>29t-m=mt</math> | <math>29t-m=mt</math> | ||
| − | <math>\frac{29t}{t+1} = | + | <math>\frac{29t}{t+1} = m</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>. | 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>. | ||
Revision as of 23:57, 9 November 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 
.
, a similar argument to the one above implies 
and 
, 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 | ||
