Mal, Num, and Pin each have distinct number of marbles. Five times the sum of
the product of the number of marbles of any two of them equals to seven times
the product of the number of the marbles the three of them have. Find the largest
possible sum of their marbles.
the question is essentially maximize $M+N+P$ with the restraint $5(MN+MP+PN)=7MNP$ and $M,N,P$ are positive integers
I tried using lagrange multipliers but it didn't seem to work, I think because the degree of the expression is 1.
the expression shows that
atleast one of the variables are a multiple of 5
hints, suggestions and solutions would all be appreciated
taken from the 2017 InImc https://chiuchang.org/wp-content/uploads/sites/2/2018/01/2017_IWYMIC_answer-2.x17381.pdf
Best Answer
Actually, we can do this question in a simple way. WLOG, let $M<N<P$.
$\dfrac{1}{M}+\dfrac{1}{N}+\dfrac{1}{P}=\dfrac{MN+MP+NP}{MNP}=\dfrac{7}{5}$
If $M=2$, then $\dfrac{1}{M}+\dfrac{1}{N}+\dfrac{1}{P}<\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{13}{12}<\dfrac{7}{5}$, therefore $M=1, \dfrac{1}{N}+\dfrac{1}{P}=\dfrac{2}{5}$
Rearrange it and we get $\left(2N-5\right)\left(2P-5\right)=25$
$\because N<P \\ \therefore \begin{cases} 2N-5=1 \\ 2P-5=25 \end{cases} \rightarrow \begin{cases} N=3 \\ P=15 \end{cases}$
$M+N+P=1+3+15=\boxed{19}$