$a+b+c+d+e=abcde$ What is $\max(a)?$ The question from my nephew is this:
If $a, b, c, d, e \in N^+$ and $a+b+c+d+e=a\times b\times c\times d\times e$, then what's the the maximum possible value of $a$? Thanks ahead:)
 A: Sorry, I've missed out one possible case. Edits are made.
By proof of contradiction, assume $1<a,b,c,d,e$, let $a=1+x_a$, $b=1+x_b$, etc.
$$5+x_a+x_b+x_c+x_d+x_e=(1+x_a)(1+x_b)(1+x_c)(1+x_d)(1+x_e)$$
By expanding RHS, you know the above equality is impossible.  So at least one number is digit $1$.
Hence,
$$5+x_a+x_b+x_c+x_d=(1+x_a)(1+x_b)(1+x_c)(1+x_d)$$
Again by expanding it, the equality is not true. Then there are two digits $1$. Eventually you may reduce that,
$$5+x_a+x_b+x_c=(1+x_a)(1+x_b)(1+x_c)$$
$$4=x_ax_b+x_ax_c+x_bx_c+x_ax_bx_c$$
Since we assume $x_a,x_b,x_c\geq0$, the two possible solutions are $(x_a,x_b,x_c)=(1,1,1),(4,1,0)$.
The maximum value of $x_a$ is $4$, so $a$ can be $5$. E.g. $a=5$, $b=2$, $c=d=e=1$.
A: $5$. This is clearly possible with $b=2$, $c=d=e=1$. It remains to show that $b+c+d+e< 6(bcde-1)$ or $6bcde>b+c+d+e+6$. However, we cannot have $b=c=d=e=1$, so $bcde\ge 2$ and, thereby, $3bcde\ge 6$. Thus we are left with $3bcde>b+c+d+e$. Assuming that $b$ is the largest number, we have $b\ge 2$, so $\frac 32bcde\ge 3cde\ge c+d+e$. Finally, $\frac32bcde>b$.
