Try small values
True or False?
For all positive integers a , ⌊ a 2 a ⌋ is even.
⌊ ⋅ ⌋ denotes the floor function .
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3 solutions
For completeness, you should show that ⌊ a 2 a ⌋ can be odd for some positive integer a .
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This might help or maybe not...
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That really helps, but wish to see the proof. However, I believe that counterexample is good enough. Maybe, the proof is not necessary to go for.
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@Michael Huang – A priori, there is no strong reason why that number should be even. If we fix a and let n vary, then for non-powers of 2, I would expect that ⌊ a 2 n ⌋ takes on odd and even values, though with (much) lower probability for odd values due to the doubling effect.
Interesting, it's always even when a is prime. Can't figure why though :P
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@Alex Li – That's a good observation.
Hint: fermat's little theorem with a bit more work.
Isn't that what I did?
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Yes. My point is that it would be interesting to show that by rigorous proof. Counterexample is the simple way to go. It would be interesting to see anyone proving the statement as they would do in the Olympiad. ;)
@Michael Huang - Even in an Olympiad, a counterexample would suffice to disprove a statement!
Even in an Olympiad, a counterexample would suffice to disprove a statement!
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I checked this MSE site out and see that counterexample might work. This site might be interesting.
I believe that when disproving the statement, giving the counterexample would be quite easy to go for. It's the way to circumvent the challenge. However, I wonder what if we were to do so during the Olympiad? Would giving counterexamples for the disproving the statement only be enough?
So much logic is going on here!
That's a nice "counterexample to the pattern established by small values".
Can principle of mathematical induction be used to conclude falsity of the statement?
I've done a proof later in this sequence showing all the possibilities to acquire a non-even answer, therefore disproving the statement (basically encompassing all counterexamples).
Wouldn't it suffice to show that it is not divisible by 2 evenly?
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That's a possible approach. How would you go about showing that?
Hang on a minute: that table's not even correct. If a=3 then 2^3/3=8/3=2.6666. It's not even an integer. I must be missing the point, because there are loads of values of 'a' for which it won't work - 3,5,6,7,9 etc. The way I'm reading the question, it will only give an integer solution if 'a' is a power of 2.
it's not asking for 2^a/a, it's asking for the floor function of 2^a/a. E.g. floor(8/3) = 2
Let o = an odd positive integer; e = an even positive integer; and n = any positive integer.
When a = o : 2^o = e (because 2^n = e as it is a series of multiplications by 2) Thus, 2^o/o = e/o which never equal an integer, so by definition cannot be even, therefore when a = any positive odd integer, the statement is false.
Thus it is false that any positive integer value of a will give an even output to f(a) = 2^a/a .
Fun problem, thank you!
No, the floor function will cut off the decimal point.
2.9378 would become 2
978.387248 would become 978
1.1 would become 1
2 stays 2 etc.
For example for n = 3 you will get 2 instead of 2+2/3.
Seem like people do not understand floor function still got this right, and I got wrong ;-;
the shit i read!!!
For a = 1 2 , ⌊ a 2 a ⌋ = 3 4 1 is odd.
This is the first counterexample in this sequence.