Difference between revisions of "2009 USAMO Problems/Problem 6"
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== Solution == | == Solution == | ||
| − | {{ | + | Suppose the <math>s_i</math> can be represented as <math>\frac{a_i}{b_i}</math> for every <math>i</math>, and suppose <math>t_i</math> can be represented as <math>\frac{c_i}{d_i}</math>. Let's start with only the first two terms in the two sequences, <math>s_1</math> and <math>s_2</math> for sequence <math>s</math> and <math>t_1</math> and <math>t_2</math> for sequence <math>t</math>. Then by the conditions of the problem, we have <math>(s_2 - s_1)(t_2 - t_1)</math> is an integer, or <math>(\frac{a_2}{b_2} - \frac{a_1}{b_1})(\frac{c_2}{d_2} - \frac{c_1}{d_1})</math> is an integer. Now we can set <math>r = \frac{b_1 b_2}{d_1 d_2}</math>, because the least common denominator of <math>s_2 - s_1</math> is <math>b_1 b_2</math> and of <math>t_2 - t_1</math> is <math>d_1 d_2</math>, and multiplying or dividing appropriately by <math>\frac{b_1 b_2}{d_1 d_2}</math> will always give an integer. |
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| + | Now suppose we kept adding <math>s_i</math> and <math>t_i</math> until we get to <math>s_m = \frac{a_m}{b_m}</math> in sequence <math>s</math> and <math>t_m = \frac{c_m}{d_m}</math> in sequence <math>t</math> so that <math>(t_m - t_i)(s_m - s_i)</math> is an integer for all <math>i</math> with <math>1 \le i < m</math>, where <math>m</math> is a positive integer. At this point, we will have <math>r</math> = <math>\frac{\prod_{n=1}^{m}b_n}{\prod_{n=1}^{m}d_n}</math>, because these are the least common denominators of the two sequences up to <math>m</math>. As we keep adding <math>s_i</math> and <math>t_i</math>, <math>r</math> will always have value <math>\frac{\prod_{n=1}^{m}b_n}{\prod_{n=1}^{m}d_n}</math>, and we are done. | ||
== See Also == | == See Also == | ||
{{USAMO newbox|year=2009|num-b=5|after=Last question}} | {{USAMO newbox|year=2009|num-b=5|after=Last question}} | ||
| − | + | ||
[[Category:Olympiad Number Theory Problems]] | [[Category:Olympiad Number Theory Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 11:19, 21 August 2020
Problem
Let 
be an infinite, nonconstant sequence of rational numbers, meaning it is not the case that 
Suppose that 
is also an infinite, nonconstant sequence of rational numbers with the property that 
is an integer for all 
and 
. Prove that there exists a rational number 
such that 
and 
are integers for all and .
Solution
Suppose the 
can be represented as 
for every , and suppose 
can be represented as 
. Let's start with only the first two terms in the two sequences, 
and 
for sequence 
and 
and 
for sequence 
. Then by the conditions of the problem, we have 
is an integer, or 
is an integer. Now we can set 
, because the least common denominator of 
is 
and of 
is 
, and multiplying or dividing appropriately by 
will always give an integer.
Now suppose we kept adding and until we get to 
in sequence and 
in sequence so that 
is an integer for all with 
, where 
is a positive integer. At this point, we will have = 
, because these are the least common denominators of the two sequences up to . As we keep adding and , will always have value , and we are done.
See Also
| 2009 USAMO (Problems • Resources) | ||
| Preceded by Problem 5 |
Followed by Last question | |
| 1 • 2 • 3 • 4 • 5 • 6 | ||
| All USAMO Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
