imqwerty
  • imqwerty
fun question (⌐■_■)
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
imqwerty
  • imqwerty
Determine all triplets (l,m,n) of positive integers such that \[l \le m \le n\]and\[l+m+n+lm+mn+nl = lmn+1\]
anonymous
  • anonymous
How is this fun lol
imqwerty
  • imqwerty
:D right nw this problem looks neat but nasty but once u see the solution u'll smile :)

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anonymous
  • anonymous
Ok we will see about that
imqwerty
  • imqwerty
sure :D
anonymous
  • anonymous
How is it that you got so many Medals and we ain't done nothing yet
Loser66
  • Loser66
@isaac4321 since it is interesting problem. Students learn a lot from it. The medal is for the Askers who have a good question also.
anonymous
  • anonymous
Are we gonna work it out
Loser66
  • Loser66
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.
imqwerty
  • imqwerty
welcome @Loser66 i'll be posting such problems time to time :) nd hey ur smart :D
dan815
  • dan815
working cases so far
dan815
  • dan815
theres no solution for all different
dan815
  • dan815
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
imqwerty
  • imqwerty
nope there r 3 sets of solutions :)
dan815
  • dan815
7,3,3 1,0,0
dan815
  • dan815
oh wait1,0,0 not possible
dan815
  • dan815
7,3,3
dan815
  • dan815
and its permutations? 3,7,3 3,3,7
imqwerty
  • imqwerty
no this question is pure algebra no permutations :)
dan815
  • dan815
|dw:1439830819873:dw|
dan815
  • dan815
well i got that solution from algebra xD
dan815
  • dan815
the other solutions have negative numbers in them
dan815
  • dan815
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
dan815
  • dan815
oh my bad i see l
dan815
  • dan815
okay lemme look for solutions in all different case too then
imqwerty
  • imqwerty
a hint - u can start it like this :) - \[l-1=p\]\[m-1=q\]\[n-1=r\] where \[p,q,r \in integers \]\[0 \le p \le q \le r\]
dan815
  • dan815
ah i tried somethjing similar with +1
dan815
  • dan815
but got lazy and said i hope theres just no solution for all different
imqwerty
  • imqwerty
:D
imqwerty
  • imqwerty
the start which i gave is jst a partial start :) u need to find a way to get through
ganeshie8
  • ganeshie8
if it helps, there are only 3 solutions |dw:1439831929718:dw|
imqwerty
  • imqwerty
yes @ganeshie8 these r the solutions :)
anonymous
  • anonymous
$$(1 + l)(1 + m)(1 + n) = 1 + l + m + n + lm + ln + mn + lmn$$ so this is equivalent to $$(1+l)(1+m)(1+n)=2(1 + lmn)$$
thomas5267
  • thomas5267
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.
dan815
  • dan815
yes oldrin
anonymous
  • anonymous
I wasn't asking for confirmation, I'm suggesting for an actual approach
dan815
  • dan815
oh ok
thomas5267
  • thomas5267
Is the following true? \[ \begin{align*} l+m+n+lm+mn+nl &= lmn+1\\ l+m+n+lm+mn+nl - lmn&=1\\ \implies \gcd(l,m,n,lm,mn,nl,lmn)=1\\ \end{align*} \] \[ \begin{align*} \gcd(l,m,n,lm,mn,nl,lmn)&=1\\ \gcd(\gcd(l,lm,nl,lmn),\gcd(m,mn),n)&=1\\ \implies \gcd(l,m,n)&=1 \end{align*} \]
dan815
  • dan815
|dw:1439833355443:dw|
anonymous
  • anonymous
yes, those follow readily from the thing I wrote above
dan815
  • dan815
ok what now?
anonymous
  • anonymous
Maybe considerations mod l,m,n
dan815
  • dan815
no just algebra its a highschool question
anonymous
  • anonymous
modular arithmetic is certainly not beyond competition high school math
dan815
  • dan815
xD im just saying for this question its pretty much all algebra
imqwerty
  • imqwerty
yes this question is all algebra :D
anonymous
  • anonymous
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
anonymous
  • anonymous
Oops , it should read \(2(1+l+m+n)=2(4+p+q+r)\) so$$pqr=2(4+p+q+r)$$
anonymous
  • anonymous
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
imqwerty
  • imqwerty
\[l-1=p\]\[m-1=q\]\[n-1=r\]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
anonymous
  • anonymous
I got lost on the 3rd number o-o
dan815
  • dan815
neat trick
dan815
  • dan815
with the bounding
dan815
  • dan815
can u show me how u got to this equation again pqr=2(p+q+r)+4
dan815
  • dan815
too much clutter to read through up there
imqwerty
  • imqwerty
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 :)
ali2x2
  • ali2x2
^ lol gg

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