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Let 8p+120 =n(n+1)/2 where n is a positive integer
16p+240 = n(n+1)

Similarly we can do
p(16p +240/p)
2p(8p+120/p)
.
.
.

From there we get p can only be 1or 2 or 3 or 5 as 240 =2*2*2*2*3*5

But when p=1 or 3 or 5 we will have no solution
So, I got p=2 only

Did I miss anything?

Can anyone tell me why 256 fails this ? It follows all conditions !

not prime

n=256 = 32*8
256 *257/ 16 -15 =4097 = 17*241 => 2 different prime numbers.

but 257 is prime, 256 should have worked ?

Something is fishy, with my hypothesis .

exactly same is the case with 448 ! :|

@sauravshakya brilliant.org?!

brilliant.org blocked me on fb! :P

And there are only two primes that satisfy this condition: \(2\) and \(47\).

Why, @shubhamsrg

I had written a program to solve this, oh well.

You're such a genius heh :-)

This problem is from Brilliant.org by the way.
Have you applied for the summer camp?

@sauravshakya
@shubhamsrg
What are your profiles on there?

Well I haven't gone on brilliant.org till now
I only liked it on fb .
Nothing else.

What will your age be in August?

I just made a account yesterday

Ah, that's bad. Students must be 13-17 for applying.

A free trip to Stanford. =)

I found some problem very interesting

That was the only reason I made account

I have done all problems except the last one.

@ParthKohli
did u get f(x)=-13+9x-3x^2+x^3
in the second last problem

No, I did a little bit of hit-and-trial. :-)

So I just counted the number of multiples of \(3\) between \(24\) and \(141\)... and voila!

Do you guys want the formalized solution anyway?

I could do better than that... but oh well.