a value of f (n)

n + sqrt(n) shud be an integer.
it will be an integer when sqrt(n) is integer.
so checking for n = 4, 9 , 16..

we get f(n) as 6,12,20...

so answer is 12 only.. D

Nice job, kvcpk!

Building on what kvcpk said, since sqrt(n) has to be an integer, that means n has to be a perfect square...

So one sneaky thing you could do is change the equation from f(n) = n + sqrt(n) to:

f(x^2) = x^2 + x

(where n = x^2 and x is a positive integer)

Factor:

f(x^2) = x(x+1)

Notice that x(x+1) denotes the product of two consecutive positive integers. It's easy to see now that the product couldn't be 5 or 10, because no two positive consecutive integers give you those values when multiplied.

But 12 = 3*4, so that works.

sanju09 wrote:If f (n) = n + √n, where n is a positive integer, which of the following would be a value of f (n)?

I. 5
II. 10
III. 12

(A) I, II and III
(B) I and III only
(C) II and III only
(D) III only
(E) None

lets take n as 4
then f(n) will be 6


let take n as 9
f(n) will be 12

here onwards value of fn will keep on increasing so only (D) III only satisy

Thats a kool trick Raz!!

Thanks for sharing!!