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}) is an integer. Now we can set < | + | 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. |
− | Now suppose we kept adding < | + | 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 == |
Latest revision as of 12:19, 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 |
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