Sum is Product

An interesting addendum to this problem is that there exists a solution for any number of positive integers being added together.

For any number n of positive integers whose sum equals their product, one can say that the numbers that work are n, 2, and (n-2) copies of 1.

For example, with 7 positive integers whose sum equals their product, the numbers 7, 2, 1, 1, 1, 1, and 1 work. 7 + 2 + 1 + 1 + 1 + 1 + 1 = 14, and 7 * 2 * 1 * 1 * 1 * 1 * 1 = 14.

$\LARGE 2 + 4 + 1 + 1 = 2\times 4 \times 1 \times 1 = 8$

The way I solved this was by starting with (P)a=S+a where P is the product of three positive integers, and S is a sum of three positive integers. I rearranged the equation to get S/(P-1)=a . Seeing this, I knew if P equaled 2, the denominator would equal 1, and that would lead to a whole number solution. I found that the only three numbers whose product is 2 is (2x1x1), so S has to be the sum of these numbers. Put this into the equation, you get 4/(2-1) and a=4. Add that to S and you get 8 .

Sorry, but I do not now Latex

We know abcd must equal a+b+c+d

From here I assumed that there would most likely be 2 1's as any more that were not 1's would make the product far exceed the sum. Let c and d both equal 1

Then:

a*b = 1 + 1 + a + b

ab = a+b+2

ab - a - b = 2

Via SFFT

(a-1)(b-1) = 3

a-1 = 1

b-1 = 3

a = 2, b = 4

1 + 1 + 2 + 4 = 8

Is there any certain pattern here.

Sum of the numbers in the equation 1=4,eqn 2=6, eqn 3=8....

Let $abcd=a+b+c+d$ . Since this equation is symmetric in $a,b,c,d$ we can let $a\ge b\ge c \ge d$ (actually this is quite a well-known technique). Now we have $abcd=a+b+c+d \le a+a+a+a=4a \implies bcd\le 4$ so $bcd\le 4$ and as $b,c,d \in \mathbb{N}$ . We can check cases. Easy to see that $(b,c,d)=(2,1,1)$ works here. Substituting back in the equation we will get that $a=4$ . So we have $(a,b,c,d)=(4,2,1,1)$ as a vaild solution. This technique is easily generalisable for $n$ numbers.

Let $abcd=a+b+c+d$ . Since this equation is symmetric in $a,b,c,d$ we can let $a\ge b\ge c \ge d$ (actually this is quite a well-known technique). Now we have $abcd=a+b+c+d \le a+a+a+a=4a \implies bcd\le 4$ so $bcd\le 4$ and as $b,c,d \in \mathbb{N}$ . We can check cases. Easy to see that $(b,c,d)=(2,1,1)$ works here. Substituting back in the equation we will get that $a=4$ . So we have $(a,b,c,d)=(4,2,1,1)$ as a vaild solution. This technique is easily generalisable for $n$ numbers.

This is great, you can go on and do 5 integers, which has the some of 10. 6 integers has the sum Of 12 and so on. Just multiply the number of integers by two and you have the answer. (5×2×1×1×1=5+2+1+1+1=10) (6×2×1×1×1×1=6+2+1+1+1+1=12) And so on...