Difference between revisions of "1989 USAMO Problems/Problem 1"
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== Problem == | == Problem == | ||
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− | <math>T_n = S_1 + S_2 + S_3 + \cdots + S_n</math> | + | For each positive [[integer]] <math>n</math>, let |
+ | <cmath> \begin{align*} | ||
+ | S_n &= 1 + \frac 12 + \frac 13 + \cdots + \frac 1n \\ | ||
+ | T_n &= S_1 + S_2 + S_3 + \cdots + S_n \\ | ||
+ | U_n &= \frac{T_1}{2} + \frac{T_2}{3} + \frac{T_3}{4} + \cdots + \frac{T_n}{n+1}. | ||
+ | \end{align*} </cmath> | ||
+ | Find, with proof, integers <math>0 < a,\ b,\ c,\ d < 1000000</math> such that <math>T_{1988} = a S_{1989} - b</math> and <math>U_{1988} = c S_{1989} - d</math>. | ||
− | + | == Solution == | |
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− | == | + | We note that for all integers <math>n \ge 2</math>, |
− | {{solution}} | + | <cmath>\begin{align*} |
+ | T_{n-1} &= 1 + \left(1 + \frac 12\right) + \left(1 + \frac 12 + \frac 13\right) + \ldots + \left(1 + \frac 12 + \frac 13 + \ldots + \frac 1{n-1}\right) \\ | ||
+ | &= \sum_{i=1}^{n-1} \left(\frac {n-i}i\right) = n\left(\sum_{i=1}^{n-1} \frac{1}{i}\right) - (n-1) = n\left(\sum_{i=1}^{n} \frac{1}{i}\right) - n \\ | ||
+ | &= n \cdot S_{n} - n . | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | It then follows that | ||
+ | <cmath>\begin{align*} | ||
+ | U_{n-1} &= \sum_{i=2}^{n} \frac{T_{i-1}}{i} = \sum_{i=2}^{n}\ (S_{i} - 1) = T_{n-1} + S_n - (n-1) - S_1 \\ | ||
+ | &= \left(nS_n - n\right) + S_n - n = (n + 1)S_n - 2n . | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | If we let <math>n=1989</math>, we see that <math>(a,b,c,d) = (1989,1989,1990, 2\cdot 1989)</math> is a suitable solution. <math>\blacksquare</math> | ||
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+ | Notice that it is also possible to use induction to prove the equations relating <math>T_n</math> and <math>U_n</math> with <math>S_n</math>. | ||
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+ | {{alternate solutions}} | ||
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+ | == See Also == | ||
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{{USAMO box|year=1989|before=First question|num-a=2}} | {{USAMO box|year=1989|before=First question|num-a=2}} | ||
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+ | * [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=356633#p356633 Discussion on AoPS/MathLinks] | ||
+ | {{MAA Notice}} | ||
[[Category:Olympiad Algebra Problems]] | [[Category:Olympiad Algebra Problems]] |
Latest revision as of 18:10, 18 July 2016
Problem
For each positive integer , let Find, with proof, integers such that and .
Solution
We note that for all integers ,
It then follows that
If we let , we see that is a suitable solution.
Notice that it is also possible to use induction to prove the equations relating and with .
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1989 USAMO (Problems • Resources) | ||
Preceded by First question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.