<s>[INST] The maximum of $a+b+c$ satisfies $2^n=a!+b!+c!$ for $n,~a,~b,~c\in\Bbb N$ How to find the maximum of $a+b+c$ satisfies $2^n=a!+b!+c!$ for $n,~a,~b,~c\in\Bbb N$? I can only deduce that $a,~b,~c$ cannot be $\geq 3$ simultaneously.
 [/INST] Wolog assume $a \le b \le c$.  So $a!|b!, c!$ so $a!|a! + b! + c!=2^n$. But as $3\not \mid 2^n$ we must have $a < 3$ so so $a = 1, 2$.
If $a = 1$ then $2\le b! +c! = 2^n-1$ and that is odd .  So $b!$ and $c!$ can't both be even.  So $b! = 1$.  And we have $1\le c! = 2^n - 2 = 2(2^n-1)$.  $2^n - 1$ is odd so only a single power of of $2|c!$ so $c \le 3$.  If $c = \color{red}1,2,3$  we can have $2 + c! = \color{red}3, 4, 8= \text{no power of two}, 2^2, 2^3$.
If $a= 2$ then $4 \le b! + c! = 2(2^n-1)$ with $2^n-1$ being odd. $b!|b! +c!$ so $b \le 3$ so $b = 2, 3$.
If $a = b=2$ then $6\le c! = 2^n -4 = 4(2^{n-1}-1)$ where $2^{n-1}-1$ odd so only $2^2$ can divide $c!$ and $2|3!$ and $8|4!$ so $c< 4$ and $c = 2,3$.  $2!+2!+2! = 6 \ne 2^n; 2! + 2! +3! = 10\ne 2^n$.
$a =2; b=3$ then $14 \le 2! + 3! + c! = 2^n$ and $c! = 2^n - 8= 8(2^{n-1}-1)$ and argument should be familiar now.  $2^3|c!$ but $2^4 \not \mid c!$ so $4 \le c \le 5$.  So $a=2; b=3; c=4,5$ so $a! + b! + c! = 2!+3!+4! = 2+6 + 24=32^5$ or $a!+b!+c! = 2!+3!+5! = 2+6 + 120=128= 2^7$.
So that's that.  $n = 7$ is the highest possible result.
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