Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

sauravshakya

  • 3 years ago

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

  • This Question is Closed
  1. UnkleRhaukus
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    *

  2. sauravshakya
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  3. sauravshakya
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 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

  4. sauravshakya
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  5. sauravshakya
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  6. sauravshakya
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  7. sauravshakya
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Did I miss anything?

  8. shubhamsrg
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 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.

  9. shubhamsrg
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  10. Jonask
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    not prime

  11. shubhamsrg
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  12. shubhamsrg
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    but 257 is prime, 256 should have worked ?

  13. shubhamsrg
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  14. shubhamsrg
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    Something is fishy, with my hypothesis .

  15. shubhamsrg
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    exactly same is the case with 448 ! :|

  16. ParthKohli
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @sauravshakya brilliant.org?!

  17. shubhamsrg
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    brilliant.org blocked me on fb! :P

  18. ParthKohli
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  19. ParthKohli
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Why, @shubhamsrg

  20. ParthKohli
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  21. shubhamsrg
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    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

  22. ParthKohli
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    You're such a genius heh :-)

  23. shubhamsrg
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 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|

  24. ParthKohli
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  25. ParthKohli
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @sauravshakya @shubhamsrg What are your profiles on there?

  26. shubhamsrg
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  27. ParthKohli
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    What will your age be in August?

  28. sauravshakya
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I just made a account yesterday

  29. ParthKohli
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  30. ParthKohli
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    A free trip to Stanford. =)

  31. sauravshakya
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I found some problem very interesting

  32. sauravshakya
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    That was the only reason I made account

  33. ParthKohli
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I have done all problems except the last one.

  34. sauravshakya
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  35. ParthKohli
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  36. ParthKohli
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 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 \(141\).

  37. ParthKohli
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  38. ParthKohli
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Do you guys want the formalized solution anyway?

  39. ParthKohli
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  40. ParthKohli
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  41. ParthKohli
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  42. mathslover
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 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.

  43. mathslover
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 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)(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 . ---**************************************---------------------------------------------------*****************************************

  44. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy