What are the possible values of these letters? Out of all the questions I answered in a math reviewer, this one killed me (and 7 more).
Let $J, K, L, M, N$ be five distinct positive integers such that
$$
\frac{1}{J} + \frac{1}{K} + \frac{1}{L} + \frac{1}{M} + \frac{1}{N} + \frac{1}{JKLMN} = 1.
$$
Then, what is $J + K + L + M + N$?
I have been thinking about this for nearly 6 days.
 A: Unless I've made a mistake, the solutions (up to permutation) are
[2, 3, 7, 43, 1807]
[2, 3, 7, 47, 395]
[2, 3, 11, 23, 31]
Maple code:
f:= proc(S) local R;
R:= map(t -> 1/t, S);
convert(R,`+`) + convert(R,`*`)
end proc:
for jj from 2 to 3 do
  for kk from jj+1 while f([jj,kk,kk+1,kk+2,kk+3]) >= 1 do
    lmin:= floor(solve(1/jj+1/kk+1/l=1));
    for ll from max(kk+1,lmin) while f([jj,kk,ll,ll+1,ll+2]) >= 1 do
       if 1/jj+1/kk+1/ll >= 1 then next fi;
       for mm from max(ll+1,floor(solve(1/jj+1/kk+1/ll+1/m=1))) while f([jj,kk,ll,mm,mm+1]) >= 1 do
          nn:= solve(f([jj,kk,ll,mm,n])=1);
          if nn::integer and nn > mm then
            printf("Found [%d, %d, %d, %d, %d]\n",jj,kk,ll,mm,nn)
          fi
 od od od od:

A: Searching through brute force gives a solution $\{2, 3, 11, 23, 31 \}$
Assume $J < K < L < M < N$ and 
also note that the least number $J$ can only be $2$ or $3$
In Python $3.x$, you can check by running this code
for j in range(2, 4):
    for k in range(j+1, 100):
        for l in range(k+1, 100):
            for m in range(l+1, 100):
                for n in range(m+1, 100):
                    if k*l*m*n + j*l*m*n + j*k*m*n + j*k*l*n + j*k*l*m + 1 == j*k*l*m*n:
                        print(j, k , l , m , n)

A: Induction could lead you to the answer. 
The equation is :
$$
    \frac 1 {x_1} + \frac 1 {x_2} + \dots + \frac 1 {x_{n}} + \frac 1 { x_1 x_2 \dots x_{n}} = 1
$$
Case $ n = 0 $: the empty set solves the equation as an empty product is 1
Case $ n = 1 $: the obvious solution is $ x_1 = 2 $.
Case $ n = 2 $: a bit more difficult, but you can find $ x_1 = 2, x_2 = 3 $.
Doing this, I noticed one thing: assuming that you solved the $ (n-1) $-th equation, you can pick $ x_n $ so that $ + \frac 1 {x_{n}} $ in the first part of the equation compensates the factor $ \frac 1 {x_n} $. Let’s check.
Case any $ n $: assuming that $ x_1, \dots x_{n} $ solves the equation, we require $ x_{n+1} $ so that
$$
    \frac 1 {x_n} + \frac 1 {x_2} + \dots + \frac 1 {x_{n+1}} + \frac 1 { x_1 x_2 \dots x_{n+1}} = \frac 1 {x_1} + \frac 1 {x_2} + \dots + \frac 1 {x_n} + \frac 1 { x_1 x_2 \dots x_n}
$$
Removing identical summands:
$$
\frac 1 {x_{n+1}} + \frac 1 { x_1 x_2 \dots x_{n+1}} = \frac 1 { x_1 x_2 \dots x_n }
$$
Multiplying tops by $ x_1 x_2 \dots x_{n+1} $ :
$$
x_1 x_2 \dots x_{n} + 1 = x_{n+1}
$$
Solved!
A: A start, on my phone.
Assume $j<k<l<m<n.$
Then j=2 or 3 because 1 makes the sum too large and 4 makes it too small.
Therefore the left without 1/j is 1/2 or 2/3.
You can get a tree of possiblities by continuing in this way.
Another tack:
Clear fractions to get
$klmn+jlmn+jkmn+jkln+jklm+1=jklmn$
or
$klmn+j(...)+1=jklmn$
or
$j(klmn-...)=klmn+1$.
Therefore $j|(klmn+1)$
(and similarly for the others)
so that j is relatively prime to the others.
Therefore all the variables are
pairwise relatively prime.
I'll leave it at this since
that's all I can think of
lying in bed.
A: $\{2,3,7,43,1807 \}$ - the first 5 terms of Sylvester's sequence - also works. In this sequence each term is the product of the previous terms plus $1$.
So it looks like the solution is not unique.
(Just saw that Robert Israel already made this observation).
