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imqwerty
 one year ago
fun question (⌐■_■)
imqwerty
 one year ago
fun question (⌐■_■)

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imqwerty
 one year ago
Best ResponseYou've already chosen the best response.9Determine all triplets (l,m,n) of positive integers such that \[l \le m \le n\]and\[l+m+n+lm+mn+nl = lmn+1\]

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.9:D right nw this problem looks neat but nasty but once u see the solution u'll smile :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok we will see about that

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How is it that you got so many Medals and we ain't done nothing yet

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1@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
 one year ago
Best ResponseYou've already chosen the best response.0Are we gonna work it out

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Thanks @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
 one year ago
Best ResponseYou've already chosen the best response.9welcome @Loser66 i'll be posting such problems time to time :) nd hey ur smart :D

dan815
 one year ago
Best ResponseYou've already chosen the best response.1theres no solution for all different

dan815
 one year ago
Best ResponseYou've already chosen the best response.1looks 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
 one year ago
Best ResponseYou've already chosen the best response.9nope there r 3 sets of solutions :)

dan815
 one year ago
Best ResponseYou've already chosen the best response.1oh wait1,0,0 not possible

dan815
 one year ago
Best ResponseYou've already chosen the best response.1and its permutations? 3,7,3 3,3,7

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.9no this question is pure algebra no permutations :)

dan815
 one year ago
Best ResponseYou've already chosen the best response.1well i got that solution from algebra xD

dan815
 one year ago
Best ResponseYou've already chosen the best response.1the other solutions have negative numbers in them

dan815
 one year ago
Best ResponseYou've already chosen the best response.1and 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
 one year ago
Best ResponseYou've already chosen the best response.1oh my bad i see l<m<n lol

dan815
 one year ago
Best ResponseYou've already chosen the best response.1okay lemme look for solutions in all different case too then

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.9a hint  u can start it like this :)  \[l1=p\]\[m1=q\]\[n1=r\] where \[p,q,r \in integers \]\[0 \le p \le q \le r\]

dan815
 one year ago
Best ResponseYou've already chosen the best response.1ah i tried somethjing similar with +1

dan815
 one year ago
Best ResponseYou've already chosen the best response.1but got lazy and said i hope theres just no solution for all different

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.9the start which i gave is jst a partial start :) u need to find a way to get through

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0if it helps, there are only 3 solutions dw:1439831929718:dw

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.9yes @ganeshie8 these r the solutions :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0$$(1 + l)(1 + m)(1 + n) = 1 + l + m + n + lm + ln + mn + lmn$$ so this is equivalent to $$(1+l)(1+m)(1+n)=2(1 + lmn)$$

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1I 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I wasn't asking for confirmation, I'm suggesting for an actual approach

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1Is the following true? \[ \begin{align*} l+m+n+lm+mn+nl &= lmn+1\\ l+m+n+lm+mn+nl  lmn&=1\\ \implies \gcd(l,m,n,lm,mn,nl,lmn)=1\\ \end{align*} \] \[ \begin{align*} \gcd(l,m,n,lm,mn,nl,lmn)&=1\\ \gcd(\gcd(l,lm,nl,lmn),\gcd(m,mn),n)&=1\\ \implies \gcd(l,m,n)&=1 \end{align*} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes, those follow readily from the thing I wrote above

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Maybe considerations mod l,m,n

dan815
 one year ago
Best ResponseYou've already chosen the best response.1no just algebra its a highschool question

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0modular arithmetic is certainly not beyond competition high school math

dan815
 one year ago
Best ResponseYou've already chosen the best response.1xD im just saying for this question its pretty much all algebra

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.9yes this question is all algebra :D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I can't really type well on my phone but I think it helps if you write $$(1l)(1m)(1n)=1(l+m+n)+lm+ln+mnlmn\\(l1)(m1)(n1)=1+l+m+n(lm+ln+mn)+lmn\\(l1)(m1)(n1)=(lmn+1)(l+m+n+lm+ln+nm)+2(lmn+l+m+n)\\(l1)(m1)(n1)=2(lmn+l+m+n)$$ Or something along those lines

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oops , it should read \(2(1+l+m+n)=2(4+p+q+r)\) so$$pqr=2(4+p+q+r)$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.9\[l1=p\]\[m1=q\]\[n1=r\]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 \[(qr)(r2)=10\]This gives \[q2=1 \],\[r2=10\] OR \[q2=\],\[r2=5\] 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 (q1)(r1)=5 hence q1=1 and r1=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
 one year ago
Best ResponseYou've already chosen the best response.0I got lost on the 3rd number oo

dan815
 one year ago
Best ResponseYou've already chosen the best response.1can u show me how u got to this equation again pqr=2(p+q+r)+4

dan815
 one year ago
Best ResponseYou've already chosen the best response.1too much clutter to read through up there

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.9i 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 :)
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