Difference between revisions of "1989 USAMO Problems/Problem 1"
(cleanup) |
(revised solution) |
||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
+ | |||
For each positive [[integer]] <math>n</math>, let | 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>. | 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 == | == Solution == | ||
− | |||
+ | We note that for all integers <math>n \ge 2</math>, | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | T_{n-1} &= | + | 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 \\ | &= \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> | + | &= 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> | |
− | |||
− | |||
− | + | {{alternate solutions}} | |
== See also == | == See also == | ||
+ | |||
{{USAMO box|year=1989|before=First question|num-a=2}} | {{USAMO box|year=1989|before=First question|num-a=2}} | ||
+ | |||
+ | * [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=356633#p356633 Discussion on AoPS/MathLinks] | ||
[[Category:Olympiad Algebra Problems]] | [[Category:Olympiad Algebra Problems]] |
Revision as of 17:57, 11 January 2008
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.
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 |