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imqwerty

  • one year ago

fun question (⌐■_■)

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  1. imqwerty
    • one year ago
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    Determine all triplets (l,m,n) of positive integers such that \[l \le m \le n\]and\[l+m+n+lm+mn+nl = lmn+1\]

  2. anonymous
    • one year ago
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    How is this fun lol

  3. imqwerty
    • one year ago
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    :D right nw this problem looks neat but nasty but once u see the solution u'll smile :)

  4. anonymous
    • one year ago
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    Ok we will see about that

  5. imqwerty
    • one year ago
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    sure :D

  6. anonymous
    • one year ago
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    How is it that you got so many Medals and we ain't done nothing yet

  7. Loser66
    • one year ago
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    @isaac4321 since it is interesting problem. Students learn a lot from it. The medal is for the Askers who have a good question also.

  8. anonymous
    • one year ago
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    Are we gonna work it out

  9. Loser66
    • one year ago
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    Thanks @imqwerty your problems inspired me a lot. They tell me how not good I am, how stupid I am so that I have to try harder on my studying.

  10. imqwerty
    • one year ago
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    welcome @Loser66 i'll be posting such problems time to time :) nd hey ur smart :D

  11. dan815
    • one year ago
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    working cases so far

  12. dan815
    • one year ago
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    theres no solution for all different

  13. dan815
    • one year ago
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    looks like theres solutions for 2 same and 1 different, i think theres an argument here for all different not being possible something intuitively tells me that

  14. imqwerty
    • one year ago
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    nope there r 3 sets of solutions :)

  15. dan815
    • one year ago
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    7,3,3 1,0,0

  16. dan815
    • one year ago
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    oh wait1,0,0 not possible

  17. dan815
    • one year ago
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    7,3,3

  18. dan815
    • one year ago
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    and its permutations? 3,7,3 3,3,7

  19. imqwerty
    • one year ago
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    no this question is pure algebra no permutations :)

  20. dan815
    • one year ago
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    |dw:1439830819873:dw|

  21. dan815
    • one year ago
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    well i got that solution from algebra xD

  22. dan815
    • one year ago
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    the other solutions have negative numbers in them

  23. dan815
    • one year ago
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    and i think theres a AM GM argument here about why all different is not possible for hte positive integers so the solution has to exist for 2 same, 1 different case only (L,m,m) once u find a solution for this u can multiple it by 3 since its m,L,m m,m,L same thing so we have to find integer solutions for this equation L+2m+2*mL +m^2=m^2L+1

  24. dan815
    • one year ago
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    oh my bad i see l<m<n lol

  25. dan815
    • one year ago
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    okay lemme look for solutions in all different case too then

  26. imqwerty
    • one year ago
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    a hint - u can start it like this :) - \[l-1=p\]\[m-1=q\]\[n-1=r\] where \[p,q,r \in integers \]\[0 \le p \le q \le r\]

  27. dan815
    • one year ago
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    ah i tried somethjing similar with +1

  28. dan815
    • one year ago
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    but got lazy and said i hope theres just no solution for all different

  29. imqwerty
    • one year ago
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    :D

  30. imqwerty
    • one year ago
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    the start which i gave is jst a partial start :) u need to find a way to get through

  31. ganeshie8
    • one year ago
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    if it helps, there are only 3 solutions |dw:1439831929718:dw|

  32. imqwerty
    • one year ago
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    yes @ganeshie8 these r the solutions :)

  33. anonymous
    • one year ago
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    $$(1 + l)(1 + m)(1 + n) = 1 + l + m + n + lm + ln + mn + lmn$$ so this is equivalent to $$(1+l)(1+m)(1+n)=2(1 + lmn)$$

  34. thomas5267
    • one year ago
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    I noticed that in each of the solution, two pairs are coprime and one isn't. For example, 2 and 8 are not coprime and (2,5) and (5,8) are both coprime.

  35. dan815
    • one year ago
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    yes oldrin

  36. anonymous
    • one year ago
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    I wasn't asking for confirmation, I'm suggesting for an actual approach

  37. dan815
    • one year ago
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    oh ok

  38. thomas5267
    • one year ago
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    Is the following true? \[ \begin{align*} l+m+n+lm+mn+nl &= lmn+1\\ l+m+n+lm+mn+nl - lmn&=1\\ \implies \gcd(l,m,n,lm,mn,nl,lmn)=1\\ \end{align*} \] \[ \begin{align*} \gcd(l,m,n,lm,mn,nl,lmn)&=1\\ \gcd(\gcd(l,lm,nl,lmn),\gcd(m,mn),n)&=1\\ \implies \gcd(l,m,n)&=1 \end{align*} \]

  39. dan815
    • one year ago
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    |dw:1439833355443:dw|

  40. anonymous
    • one year ago
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    yes, those follow readily from the thing I wrote above

  41. dan815
    • one year ago
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    ok what now?

  42. anonymous
    • one year ago
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    Maybe considerations mod l,m,n

  43. dan815
    • one year ago
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    no just algebra its a highschool question

  44. anonymous
    • one year ago
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    modular arithmetic is certainly not beyond competition high school math

  45. dan815
    • one year ago
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    xD im just saying for this question its pretty much all algebra

  46. imqwerty
    • one year ago
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    yes this question is all algebra :D

  47. anonymous
    • one year ago
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    I can't really type well on my phone but I think it helps if you write $$(1-l)(1-m)(1-n)=1-(l+m+n)+lm+ln+mn-lmn\\(l-1)(m-1)(n-1)=-1+l+m+n-(lm+ln+mn)+lmn\\(l-1)(m-1)(n-1)=(lmn+1)-(l+m+n+lm+ln+nm)+2(lmn+l+m+n)\\(l-1)(m-1)(n-1)=2(lmn+l+m+n)$$ Or something along those lines

  48. anonymous
    • one year ago
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    Oops , it should read \(2(1+l+m+n)=2(4+p+q+r)\) so$$pqr=2(4+p+q+r)$$

  49. anonymous
    • one year ago
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    I probably made a few typos typing on my phone since OpenStudy lags so much and I can't even see what I'm typing but something like thet

  50. imqwerty
    • one year ago
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    \[l-1=p\]\[m-1=q\]\[n-1=r\]so the equation can be written in the form\[pqr=2(p+q+r)+4\] where p,q,r are integers such that\[0 \le p \le q \le r\]. we can clearly see that p=0 is not possible so we can write the equation-\[pqr=2(p+q+r)+4\] as\[2(1\div pq + 1\div qr + 1\div rp) + 4\div pqr = 1\]. IF \[p \ge 3\] then \[q \ge 3 \]and\[r \ge 3\] so the left side is bounded by 6/9+4/27 which is less than 1. We conclude that p=1 or 2. CASE 1:- suppose p=1 then we have \[qr=2(q+r)+6\]or \[(q-r)(r-2)=10\]This gives \[q-2=1 \],\[r-2=10\] OR \[q-2=\],\[r-2=5\] recall\[q \le r\]So this implies (p,q,r)=(1,3,12) or (1,4,7). CASE 2- if p=2 the equation reduces to (q-1)(r-1)=5 hence q-1=1 and r-1=5 is the only solution. this gives (p,q,r)=(2,2,6) reverting back to l,m,n finall :D we get 3 triplets - (2,4,13), (2,5,8) and (3,3,7) :D

  51. anonymous
    • one year ago
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    I got lost on the 3rd number o-o

  52. dan815
    • one year ago
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    neat trick

  53. dan815
    • one year ago
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    with the bounding

  54. dan815
    • one year ago
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    can u show me how u got to this equation again pqr=2(p+q+r)+4

  55. dan815
    • one year ago
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    too much clutter to read through up there

  56. imqwerty
    • one year ago
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    i just substituted the values of l as p+1 nd similarly m nd n nd then solved the equation nd finally the result is pqr=2(p+r+r)+4 :)

  57. ali2x2
    • one year ago
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    ^ lol gg

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