Difference between revisions of "2002 AIME I Problems/Problem 4"
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== Solution 2 == | == Solution 2 == | ||
− | Note that <math>a_1 + a_2 + \cdots + a_i = \dfrac{i}{i+1}</math>. This can be proven by induction. Thus, <math>\sum\limits_{i=m}^{n-1} a_i = \sum\limits_{i=1}^{n-1} a_i - \sum\limits_{i=1}^{m-1} a_i = \dfrac{n-1}{n} - \dfrac{m-1}{m} = \dfrac{n-m}{mn} = 1/29</math>. Cross-multiplying yields <math>29n - 29m - mn = 0</math>, and adding <math>29^2</math> to both sides gives <math>(29-m)(29+n) = 29^2</math>. Clearly, | + | Note that <math>a_1 + a_2 + \cdots + a_i = \dfrac{i}{i+1}</math>. This can be proven by induction. Thus, <math>\sum\limits_{i=m}^{n-1} a_i = \sum\limits_{i=1}^{n-1} a_i - \sum\limits_{i=1}^{m-1} a_i = \dfrac{n-1}{n} - \dfrac{m-1}{m} = \dfrac{n-m}{mn} = 1/29</math>. Cross-multiplying yields <math>29n - 29m - mn = 0</math>, and adding <math>29^2</math> to both sides gives <math>(29-m)(29+n) = 29^2</math>. Clearly, <math>m < n \implies 29 - m = 1</math> and <math>29 + n = 29^2</math>. Hence, <math>m = 28</math>, <math>n = 812</math>, and <math>m+n = \fbox{840}</math>. |
+ | |||
+ | <math>~</math> keeper1098 | ||
== 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}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 19:12, 4 May 2020
Contents
Problem
Consider the sequence defined by for . Given that , for positive integers and with , find .
Solution 1
. Thus,
Which means that
Since we need a factor of 29 in the denominator, we let .* Substituting, we get
so
Since is an integer, , so . It quickly follows that and , so .
*If , a similar argument to the one above implies and , which implies . This is impossible since .
Solution 2
Note that . This can be proven by induction. Thus, . Cross-multiplying yields , and adding to both sides gives . Clearly, and . Hence, , , and .
keeper1098
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 |
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