Bmo1 q4, 1996

Here is the question:

For any real number x, let [x] denote the greatest integer which is less than or equal to x. Define q(n) = [n/[sqrt(n)]] for n = 1,2,3,... . Determine all positive integers n for which q(n) > q(n + 1)

I think I have solved the problem but I am unable to find solutions to this anywhere. This is the way I did it:

Notice that the only way [sqrt(n)] is different from [sqrt(n+1)] is when n is 1 less than a perfect square, e.g. 8. Further, notice that the denominators would be different whilst the numerator would be the same, resulting in different solutions when rounded. Hence the answer is:

{n where n = (a^2) - 1, where a^2 is a perfect square}

Please could anyone let me know whether it is correct? Thanks.

Reply 1

mqb2766

21

Original post by entrepreneur_13

Here is the question:

For any real number x, let [x] denote the greatest integer which is less than or equal to x. Define q(n) = [n/[sqrt(n)]] for n = 1,2,3,... . Determine all positive integers n for which q(n) > q(n + 1)

I think I have solved the problem but I am unable to find solutions to this anywhere. This is the way I did it:

Notice that the only way [sqrt(n)] is different from [sqrt(n+1)] is when n is 1 less than a perfect square, e.g. 8. Further, notice that the denominators would be different whilst the numerator would be the same, resulting in different solutions when rounded. Hence the answer is:

{n where n = (a^2) - 1, where a^2 is a perfect square}

Please could anyone let me know whether it is correct? Thanks.

I thought you had Gardiners book which had some advice/hints about this paper/question?

Reply 2

entrepreneur_13

OP

10

Yeah. But I don't understand his hints. I sort of understand that he took the same approach as me with stating n as 1 less than a square, but I don't understand the rest. So just wanted to check whether my solution is right as I got that it is not a finite set of numbers.

(edited 1 year ago)

Reply 3

mqb2766

21

Original post by entrepreneur_13

Yeah. But I don't understand his hints. I sort of understand that he took the same approach as me with stating n as 1 less than a square, but I don't understand the rest. So just wanted to check whether my solution is right as I got that it is not a finite set of numbers.

The answer is correct, but your explanation is unlikely to get that many marks.
When you say "notice", how is that a proof? You have to establish properties about the sequence and show your "guessed" solution is actually the only solution that satisfies the given property. Id guess it could be done in a few ways, but Gardiners argument has a decent structure so it would be worth trying to formulate an argument like his, or if you can think of how to improve it, go for it.

But at the very least you need to show clearly that the inequality is satified when n=k^2-1 and not satisfied otherwise. This pretty much implies your proof has some inequalities in it.