Difference between revisions of "1994 USAMO Problems/Problem 1"
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+ | ==Problem== | ||
Let <math> \, k_1 < k_2 < k_3 <\cdots\, </math>, be positive integers, no two consecutive, and let <math> \, s_m = k_1+k_2+\cdots+k_m\, </math>, for <math> \, m = 1,2,3,\ldots\;\; </math>. Prove that, for each positive integer <math>n</math>, the interval <math> \, [s_n, s_{n+1})\, </math>, contains at least one perfect square. | Let <math> \, k_1 < k_2 < k_3 <\cdots\, </math>, be positive integers, no two consecutive, and let <math> \, s_m = k_1+k_2+\cdots+k_m\, </math>, for <math> \, m = 1,2,3,\ldots\;\; </math>. Prove that, for each positive integer <math>n</math>, the interval <math> \, [s_n, s_{n+1})\, </math>, contains at least one perfect square. | ||
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So, <math>k_{n+1}\geq d(s_n)</math> and all intervals between <math>s_n</math> and <math>s_{n+1}</math> will contain at least one perfect square. | So, <math>k_{n+1}\geq d(s_n)</math> and all intervals between <math>s_n</math> and <math>s_{n+1}</math> will contain at least one perfect square. | ||
+ | ==See Also== | ||
+ | {{USAMO box|year=1994|before=First Problem|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | [[Category:Olympiad Number Theory Problems]] |
Revision as of 07:59, 19 July 2016
Problem
Let , be positive integers, no two consecutive, and let
, for
. Prove that, for each positive integer
, the interval
, contains at least one perfect square.
Solution
We want to show that the distance between and
is greater than the distance between
and the next perfect square following
.
Given , where no
are consecutive, we can put a lower bound on
. This occurs when all
:
Rearranging, . So,
, and the distance between
and
is
.
Also, let be the distance between
and the next perfect square following
. Let's look at the function
for all positive integers
.
When is a perfect square, it is easy to see that
.
Proof: Choose
.
.
When is not a perfect square,
.
Proof: Choose
with
.
.
So, for all
and
for all
.
Now, it suffices to show that for all
.
So, and all intervals between
and
will contain at least one perfect square.
See Also
1994 USAMO (Problems • Resources) | ||
Preceded by First Problem |
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.