Inspired by Nihar Mahajan's featured problem

Let $a\leq b\leq c\leq d\leq e$ be positive integers such that $a+b+c+d+e = abcde.$ Then $abcde = a+b+c+d+e \leq 5e \implies abcd \leq 5$

There are only a few ways to have the product of four positive integers be at most $5$ :

• $(a,b,c,d) = (1,1,1,1)$ : Then the equation $a+b+c+d+e = abcde$ becomes $4+e = e$ which has no solution.
• $(a,b,c,d) = (1,1,1,2)$ : Then the equation becomes $5+e = 2e \implies e=5$
• $(a,b,c,d) = (1,1,1,3)$ : Then the equation becomes $6+e = 3e \implies e=3$
• $(a,b,c,d) = (1,1,1,4)$ : Then the equation becomes $7+e = 4e \implies e=\frac{7}{3},$ which isn't an integer.
• $(a,b,c,d) = (1,1,2,2)$ : Then the equation becomes $6+e = 4e \implies e=2$
• $(a,b,c,d) = (1,1,1,5)$ : Then the equation becomes $8+e = 5e \implies e=2$ which fails $d \leq e$ (it's a repeat)

It follows there are three possibilities: $(a,b,c,d,e) \in \Big\{(1,1,1,2,5), (1,1,1,3,3), (1,1,2,2,2)\Big\}$ which gives an answer of $(1\times 1\times 1\times 2\times 5) + (1\times 1\times 1\times 3\times 3) + (1\times 1\times 2\times 2\times 2) = 10 + 9 + 8 = \boxed{27}$

• Solutions of the form $(1,1,1,1,x)$ don't exist, since $x = 4+x$ cannot be solved.

• Solutions of the form $(1,1,1,x,y)$ are possible. We get $\begin{aligned} xy & = 3 + x + y \\ (x - 1)(y - 1) & = 4 \\ \{x,y\} & = \{3,3\}\ \text{or}\ \{2,5\}.\end{aligned}$ This results in the solutions $1 + 1 + 1 + 3 + 3 = 1 \times 1 \times 1 \times 3 \times 3 = \boxed{9}$ , and $1 + 1 + 1 + 2 + 5 = 1 \times 1 \times 1 \times 2 \times 5 = \boxed{10}$ .

• Solutions of the form $(1,1,x,y,z)$ : write $x = 2 + \alpha$ , $y = 2 + \beta$ , $z = 2 + \gamma$ ; of course, $\alpha,\beta,\gamma \geq 0$ . Then $xyz = (2+\alpha)(2+\beta)(2+\gamma) = 8 + 4(\alpha + \beta + \gamma) + \cdots$ while $2+x+y+z = 8 + \alpha + \beta + \gamma.$ The only way in which these can be equal if $\alpha + \beta + \gamma = 0$ , i.e. they are all zero, so that $x = y = z = 2$ . This gives the solution $1 + 1 + 2 + 2 + 2 = 1 \times 1 \times 2 \times 2 \times 2 = \boxed{8}$ .

• Solutions with fewer than two 1s are easily ruled out.

$\large{1+1+1+2+5=1\times1\times1\times2\times5}$

$\large{1+1+1+3+3=1\times1\times1\times3\times3}$

$\large{1+1+2+2+2=1\times1\times2\times2\times2}$

The required sum is $10+9+8=\boxed{27}$ .

For who likes to write code, I tried to solve it with Python. Since the higher numbers we try, the bigger is the gap between the sum and the product, I set a range of 100 numbers to try. The shell stops after a while, just cut off redundat solutions (permutations of the same numbers) and you will get as solutions 10, 9 and 8. So 10+9+8 = 27. The code I used is this:

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for a in range(101):
    for b in range(101):
        for c in range(101):
            for d in range(101):
                for e in range(101):
                    if a+b+c+d+e== a*b*c*d*e:
                        print(f'The solution is', a, b, c, d, e)
                        print(a+b+c+d+e) 

There are three solutions with n=5, (1,1,1,2,5), (1,1,1,3,3) and (1,1,2,2,2) with equal sum/product. Answer=10+9+8=27.

Here's an exhaustive search in Python 2.7.10.

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import itertools
def product_fun(a_list):
   n=1
      for i in a_list:
        n=n*i
   return n

lst=[1,2,3,4,5]
summed_it=0
for x in itertools.combinations_with_replacement(lst,5):
  for i in itertools.permutations(x,5):
     if sum(i)==product_fun(i):
       print "Here is one: ",i
       print "sum ",sum(i)," product", product_fun(i)
       print "--------"
       summed_it(i)+=sum(i)
       break
print "The sum of sums and/or products: ",summed_it 

I did MMC for the numbers between 5 to 15 (just for sure), than sum the values and multiplies to check if that number is a solution