Difference between revisions of "1997 JBMO Problems/Problem 5"
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Since it’s impossible to have exactly one or zero even numbers, at least two of the integers must be even. | Since it’s impossible to have exactly one or zero even numbers, at least two of the integers must be even. | ||
− | == See | + | == See Also == |
{{JBMO box|year=1997|num-b=4|after=Last Problem}} | {{JBMO box|year=1997|num-b=4|after=Last Problem}} | ||
[[Category:Olympiad Number Theory Problems]] | [[Category:Olympiad Number Theory Problems]] |
Latest revision as of 21:12, 7 August 2018
Problem
Let , , , be positive integers such that Show that at least two of the numbers are even.
Solution
In order to prove that at least two numbers are even, we need to prove that it is impossible to have exactly one number even and that it is impossible to have exactly no numbers even.
Lemma 1: Impossible to have exactly 1 number even
Assume one of the numbers is even. If is even, then the left side is congruent to modulo while the right side is congruent to modulo which can not happen. If one of is even, then the left side is congruent to modulo while the right side is congruent to modulo which can not happen. Thus, it is impossible for exactly one of the numbers to be even.
Lemma 2: Impossible to have all numbers odd
Assume all numbers are odd (or that there are no even numbers). If is odd for then That means the left side is congruent to modulo while the right side is congruent to modulo Since this can not happen, it is impossible for none of the numbers to be even.
Since it’s impossible to have exactly one or zero even numbers, at least two of the integers must be even.
See Also
1997 JBMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 | ||
All JBMO Problems and Solutions |