Question

Find all prime numbers "p", such that 8p+120 is a triangle number?
I got only 2

Answer 1

*

Answer 2

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

Answer 3

Now,
2(8p+120) from here no solution as 2+1<120
4(4p+60) from here no solution as 4+1<60
8(2p+30) from here no solution as 8+1<30
16(p+15) here we have a solution when p=2 an i.e 16*17

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

Did I miss anything?

Answer 8

16p + 240 = n(n+1)
p = n(n+1)/16 - 15 = n(n+1)/16 - (3*5)

n(n+1) should be divisible by 16, it should not be odd, also , it should not have 3 or 5 as factor.

So, n or n+1 should be prime, one is prime then the other should not be divisible by 3 or 5,n(n+1) should be divisible by 16, n(n+1)/16 should be even. => n(n+1) should be divisible by 32 , given 2 is excluded as our prime.

n=16 fits
31 fits


I see no other number till now, p=2 and p=47 seems legit at the moment.

Answer 9

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

Answer 10

not prime

Answer 11

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

Answer 12

but 257 is prime, 256 should have worked ?

Answer 13

352 also doesn't work, when simplified with n=352,we get 7751, which is 23*337, again a product of 2 primes !

Answer 14

Something is fishy, with my hypothesis .

Answer 15

exactly same is the case with 448 ! :|

Answer 16

@sauravshakya brilliant.org?!

Answer 17

brilliant.org blocked me on fb! :P

Answer 18

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

Answer 19

Why, @shubhamsrg

Answer 20

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

Answer 21

They had asked something, I posted a very neat and good solution in the comments, explaining it pretty well. After 2-3 days, when I checked back, They had deleted my comment! :O
I abused them! :P
And now the consequences! :P

Answer 22

You're such a genius heh :-)

Answer 23

Hell no! It was one of those rare moments when I was able to do some question. Hence I got too carried away when I came to know it wasn't there. B|

Answer 24

This problem is from Brilliant.org by the way.

Have you applied for the summer camp?

Answer 25

@sauravshakya
@shubhamsrg

What are your profiles on there?

Answer 26

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

Answer 27

What will your age be in August?

Answer 28

I just made a account yesterday

Answer 29

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

Answer 30

A free trip to Stanford. =)

Answer 31

I found some problem very interesting

Answer 32

That was the only reason I made account

Answer 33

I have done all problems except the last one.

Answer 34

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

Answer 35

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

Answer 36

The maximum number that you could get from the set was \(44 +47 + 50 =141\) and the least was \(24\).

Then I just randomly added some set elements and I observed that you always get a multiple of \(3\) between \(24\) and \(141\).

Answer 37

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

Answer 38

Do you guys want the formalized solution anyway?

Answer 39

All the set elements are \(2\) modulo \(3\), and adding three numbers that are \(2\) modulo \(3\) gives us a number divisible by \(3\).

Answer 40

We can simply assert that all sums will be divisible by \(3\), where the maximum is \(141\) and the minimum is \(24\).

Answer 41

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

Answer 42

hey I had submitted a solution for this question @sauravshakya to brilliant.org let us see what happens. by the way it is a great site.

Answer 43

Go for hit and trial method first.
We have the first prime number as 2 : let us take p = 2 and see what we get : 8p + 120 = 8(2) + 120 = 16+120 = 136.
Is 136 in the form of n(n+1)/2 ? Let us check it .
n(n+1)/2 = 136
n(n+1) = 272
n^2 + n - 272 = 0
from this we get n = 16 and so we get : 16 * 17/2 = 136. That is 2 is our first prime number that satisfies the given equation : 8p + 120 . See, we know that a triangular number forms an equilateral triangle . i.e. 8p + 120 will form an equilateral triangle. We know that if n is a triangular number then 8n + 1 is a square ( find the proof here : http://nrich.maths.org/790&part=solution ) therefore we have : 8(8p + 120) + 1 is a square 64 p + 961 will be a square. Let the side of that square be a therefore we have : 64 p + 961 = a^2 a^2 - 961 = 64 p a^2 - 31^2 = 64p (a+31)(a-31) = 64p .
We have : (a+31)(a-31) = 64p
Let us first take a as odd
Since a is odd therefore it must be in the form of 2k+1 ( k is even )
Therefore we have : (2k + 32)(2k – 30) = 64p
(k + 16)(k – 15) = 16p
Since k is even therefore it can be written in the form of \(2\lambda\)
\((2\lambda + 16)(2\lambda – 15) = 16p\)
\((\lambda+8) (2\lambda – 15) = 8p\)
Therefore \(8|\lambda +8\) since 8 divides 8 and so it must divide also
Therefore \(8| \lambda\) or \(16|2\lambda\) or \(16|k\) (We have taken earlier as k = \(2\lambda\))
Since 16 divides k so we can write k as \(16\alpha\) .
Putting this in the equation : (2k+32)(2k-30) = 64p
\((2(16\alpha)+32)(2(16\alpha)-30) = 64p\)
\((32\alpha+64)(32\alpha -30) = 64p\)
\((\alpha+1)(16\alpha-15)=p\)
Oh! Here we have written a prime number ( p ) in the form of two composite numbers multiplication.
But we have done everything right so there must be like : \(16\alpha – 15 = 1\) and \(\alpha + 1 = p\)
Therefore \(alpha = 1\) and p = 2 which we have already found. This means k is not in the form of \(2\lambda\)
Let us take it now as \(k = 2\delta + 1\)
We have : \((k+16)(k-15) = 16p\) - this we got earlier
\((2\delta+1 + 16)(2\delta +1 – 15) = 16p\)
\((2\delta+17)(2\delta – 14) = 16p\)
\((2\delta+17)(\delta – 7) = 8p\)
Since \(2\delta + 17\) is odd ( since \(2\delta\) is even and even + odd = odd ) therefore
8 must divide \(\delta-7\) . Let us take \(\delta – 7 \) = \(8 \beta \)
\(\delta = 8\beta + 7\)
Since k = \(2\delta + 1\) therefore we have :
\(k = 2(8\beta +7)+1\) and thus we get :
\((k+16)(k-15) = 16p\) as :
\((16\beta + 31)(16\beta) = 16p\)
\((16\beta + 31)(\beta) = p\)
Again we have p as the multiplication of two composite numbers . So it must be like \(\beta= 1\) and \(16\beta+31 = p\)
Therefore \(\beta = 1\) and \(16\beta + 31 = 16(1) + 31 = 16+31 = 47 = p \)
Therefore another prime number is 47 and so the sum of the prime numbers is 49 .
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