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

This Question is Closed

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

sauravshakya
 one year ago
Best ResponseYou've already chosen the best response.1Now, 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

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

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

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

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

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.316p + 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.

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

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

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

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.3352 also doesn't work, when simplified with n=352,we get 7751, which is 23*337, again a product of 2 primes !

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

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

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

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

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

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

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.3They 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

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

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.3Hell 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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

mathslover
 one year ago
Best ResponseYou've already chosen the best response.2hey 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.

mathslover
 one year ago
Best ResponseYou've already chosen the best response.2Go 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 . *******************************************************************************
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