Question

Theory number.
n . (2^n) is perfect square number with n is elements of set {1 , 2, 3, 4, ... , 2015}. How many n which satisfy this condition.

Answer 1

looks n is even number , right ?

Answer 2

Yes, n would have to be an even number. That is necessary, but not sufficient.

Answer 3

I think n also have to be a perfect square itself.

Answer 4

The odd factors definitely have to be squared.

Answer 5

Let me think about the factors of 2 in n.

Answer 6

n = 2^k with k = 2,4,6, ...

Answer 7

Having zero powers of 2 does not work.
Having one power of two does not work.
Having two powers of two does work.

Answer 8

No, n = p^2 * 2^k with k = 2, 4, 6...

where p is an odd number.

Answer 9

So let's take k = 2 first,

2^2 = 4
2015/4 = 503.75
Then we want how many odd perfect squares less than 503.75
sqrt(503.75) = 22.44
So that looks like we will have odd squares for 1 through 21

That's 11, yes?

Answer 10

looks yours is same with mine
n . 2^n = p^2
for n an even number so 2^n is already be perfect, therefore n also must be perfect square. because perfect square times perfect square is perfect square.