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

This Question is Closed

sauravshakyaBest ResponseYou've already chosen the best response.1
Let 8p+120 =n(n+1)/2 where n is a positive integer 16p+240 = n(n+1)
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
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
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
Similarly we can do p(16p +240/p) 2p(8p+120/p) . . .
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
From there we get p can only be 1or 2 or 3 or 5 as 240 =2*2*2*2*3*5
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
But when p=1 or 3 or 5 we will have no solution So, I got p=2 only
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
Did I miss anything?
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.3
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.
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.3
Can anyone tell me why 256 fails this ? It follows all conditions !
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.3
n=256 = 32*8 256 *257/ 16 15 =4097 = 17*241 => 2 different prime numbers.
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.3
but 257 is prime, 256 should have worked ?
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.3
352 also doesn't work, when simplified with n=352,we get 7751, which is 23*337, again a product of 2 primes !
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.3
Something is fishy, with my hypothesis .
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.3
exactly same is the case with 448 ! :
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
@sauravshakya brilliant.org?!
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.3
brilliant.org blocked me on fb! :P
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
And there are only two primes that satisfy this condition: \(2\) and \(47\).
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
I had written a program to solve this, oh well.
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.3
They had asked something, I posted a very neat and good solution in the comments, explaining it pretty well. After 23 days, when I checked back, They had deleted my comment! :O I abused them! :P And now the consequences! :P
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
You're such a genius heh :)
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.3
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
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
This problem is from Brilliant.org by the way. Have you applied for the summer camp?
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
@sauravshakya @shubhamsrg What are your profiles on there?
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.3
Well I haven't gone on brilliant.org till now I only liked it on fb . Nothing else.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
What will your age be in August?
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
I just made a account yesterday
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Ah, that's bad. Students must be 1317 for applying.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
A free trip to Stanford. =)
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
I found some problem very interesting
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
That was the only reason I made account
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
I have done all problems except the last one.
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
@ParthKohli did u get f(x)=13+9x3x^2+x^3 in the second last problem
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
No, I did a little bit of hitandtrial. :)
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
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 \(141\).
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
So I just counted the number of multiples of \(3\) between \(24\) and \(141\)... and voila!
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Do you guys want the formalized solution anyway?
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
All the set elements are \(2\) modulo \(3\), and adding three numbers that are \(2\) modulo \(3\) gives us a number divisible by \(3\).
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
We can simply assert that all sums will be divisible by \(3\), where the maximum is \(141\) and the minimum is \(24\).
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
I could do better than that... but oh well.
 one year ago

mathsloverBest ResponseYou've already chosen the best response.2
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.
 one year ago

mathsloverBest ResponseYou've already chosen the best response.2
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)(a31) = 64p . We have : (a+31)(a31) = 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 \(162\lambda\) or \(16k\) (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)(2k30) = 64p \((2(16\alpha)+32)(2(16\alpha)30) = 64p\) \((32\alpha+64)(32\alpha 30) = 64p\) \((\alpha+1)(16\alpha15)=p\) 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 : \((k+16)(k15) = 16p\)  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 \(\delta7\) . 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 : \((k+16)(k15) = 16p\) 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 . *******************************************************************************
 one year ago
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