Five Easy Pieces

Without loss of generality assume that $0 \lt a \le b \le c \le d \le e$ . Then $a + b + c + d + e \le 5e$ .

Now suppose $abcd \gt 5$ . Then since $e$ is positive we have that $abcde \gt 5e$ , which combined with the previous result implies that $abcde \gt 5e \ge a + b + c + d + e$ . So since we wish to have $abcde = a + b + c + d + e$ we are forced to conclude that $abcd \le 5$ . This leaves us with the following cases to examine:

• $abcd = 1 \Longrightarrow e = a + b + c + d + e \Longrightarrow a + b + c + d = 0$ , which is not valid as $a,b,c,d \gt 0$ .

• $abcd = 2 \Longrightarrow 2e = a + b + c + d + e \Longrightarrow e = a + b + c + d.$ Now as $abcd = 2$ and $0 \lt a \le b \le c \le d$ we must have $(a,b,c,d) = (1,1,1,2)$ , which yields $e = 5$ and $abcde = a + b + c + d + e = 10$ .

• $abcd = 3 \Longrightarrow 3e = a + b + c + d + e \Longrightarrow 2e = a + b + c + d$ . Now as $abcd = 3$ we must have that $(a,b,c,d) = (1,1,1,3)$ , which yields $2e = 6 \Longrightarrow e = 3$ , which in turn yields that $abcde = a + b + c + d + e = 9$ .

• $abcd = 4 \Longrightarrow 4e = a + b + c + d + e \Longrightarrow 3e = a + b + c + d$ . Now as $abcd = 4$ we can have either $(a,b,c,d) = (1,1,1,4)$ or $(a,b,c,d) = (1,1,2,2)$ . In the first case we find that $e$ is non-integral, and in the second case that $3e = 6 \Longrightarrow e = 2$ , which yields $abcde = a + b + c + d + e = 8$ .

• $abcd = 5 \Longrightarrow 5e = a + b + c + d + e \Longrightarrow 4e = a + b + c + d$ . Now as $abcd = 5$ we must have that $(a,b,c,d) = (1,1,1,5)$ , which yields $e = 2$ , in which case $d \gt e$ , and besides, simply duplicates the value for $x$ found in the $abcd = 2$ case.

So we have $n = 3$ possible values for $x$ which sum to $S = 10 + 9 + 8 = 27$ , and so $n + S = \boxed{30}$ .

Without loss of generality, suppose that $a \geq b \geq c \geq d \geq e$ .

We obtain the following system of equations:

$a(bcde-1)=b+c+d+e$

$b(acde-1)=a+c+d+e$

$c(abde-1)=a+b+d+e$

$d(abce-1)=a+b+c+e$

$e(abcd-1)=a+b+c+d$

Adding all five equations up give:

$pa+qb+rc+sd+te=4(a+b+c+d+e)$ , where $p \leq q \leq r \leq s \leq t$

We conclude that $d=e=1$ . Otherwise, $2.2.2.1-1 =7 \leq p \leq q \leq r \leq s \leq t$ , making the sum on L.H.S. greater than the sum on R.H.S., which is impossible.

Therefore, write $abc=x$ , $a+b+c=x-2$ , from which we obtain $abc=a+b+c+2$ .

Rewrite the equation as $(a-2)(bc-1)+b(c-1)+c(b-1)=4$ . Let's first suppose that $a\geq3$ .

Note that all the terms are non-negative integers. Let's first suppose that one or more of them equal $0$ , which is only possible if either $bc=1, c=1$ or $b=1$ .

However, note that $bc=1 \rightarrow b=1,c=1$ , which makes the three terms add up to $0\neq4$ .

Suppose now that $c=1$ . We obtain the equation $(a-1)(b-1)=4$ , of which possible solutions are $a=5,b=2$ , $a=3,b=3$ and $a=2,b=5$ . We reject the third solution since $a<b$ , which violates our initial assumption.

If $b=1$ , we obtain the equation $(a-1)(c-1)=4$ , but all of the solutions $(a,b,c)=(5,1,2),(3,1,3),(2,1,5)$ violates our initial assumptions.

Now, having considered the possibilities that one or more of the terms equal $0$ , we focus on cases where they are all positive.

The only partition of $4$ as three positive integers is $2+1+1$ .

Suppose that $(a-2)(bc-1)=1$ . This implies $a=3,bc=2$ , which means one of $b$ or $c$ must equal $1$ . However, this makes one of $b(c-1)$ or $c(b-1)$ equals $0$ .

Suppose that $b(c-1)=1$ . This implies that $b=1,c=2$ . However, this makes $c(b-1)=0$ . A similar argument applies when $c(b-1)=1$ .

Therefore, it is impossible that all three terms are positive.

Now, we must remember to consider separately the cases where $a\leq2$ since we suppose $a\geq3$ at the beginning. Out of all possible combinations $(2,2,2,1,1),(2,2,1,1,1),(2,1,1,1,1)$ and $(1,1,1,1,1)$ , only $(2,2,2,1,1)$ fits the criteria.

Removing our initial assumption of $a \geq b \geq c \geq d \geq e$ , we find that the all the solutions are permutations of the following:

$(2,2,2,1,1),(3,3,1,1,1),(5,2,1,1,1)$

Therefore, there are $3$ possible sums, which adds up to $8+9+10=27$ , hence the answer $3+27=\boxed{30}$

Clearly, 5 1's will not work since 1 .ne.5. If we have 4 1's, then n,1,1,1,1 are the numbers with product n and sum n + 4, so also impossible. If we have 3 1's, the numbers are n,k,1,1,1 with product nk and sum n + k + 3. Brief trial and error shows that n,k = 5,2 giving product = 10 and sum = 10. Second solution is 3,3 giving product and sum = 9. Lastly, for 2 1's, the only solution is 2,2,2,1,1 with product and sum = 8. So n = 3, the values of x are 8,9, and 10, so S = 27, and n + S = 30.