A number theory problem by Tan Peng

WLOG $a\ \leq b \leq c$ . Note that $d > c$ since factorials is a monotonicaly increasing function. We have $\begin{aligned} d! &= a!+b!+c!\\ &\leq 3c!\\ \dfrac{d!}{c!} &\leq 3\\ \dfrac{d!}{c!} &= 2, 3\\ \implies (d,c) &= (3,2), (2,1), (2,0) \end{aligned}$

We note that the $d=2$ cannot work since $n! \geq 1$ . Thus, the only possible solution is $d=3$ , $c=2$ , which yields $a=b=2$ . Thus, the only solution is $(2,2,2,3)$ .

d < 4 since 4! = 24 and the maximum value of the left side would be 3 * 3! = 18. d must also be more than 2 since 2! = 2 and the minimum of the left side is 3 * 1! = 3. Therefor d = 3 and a! + b! + c! = 3! = 6. The only solution is now 2! + 2! + 2! = 3!.

Since { $\alpha,\beta,\gamma,\delta$ }are all positive integers.Then $\alpha!,\beta!,\gamma!\,\delta!$ are all positive and { $\alpha!,\beta!,\gamma!$ } $<\delta!$
Let $\alpha!<\beta!<\gamma!<\delta!$ $\gamma!(\alpha)(\alpha-1)(\alpha-2)...(\alpha-\gamma+1)(\alpha-\gamma)+(\beta)(\beta-1)(\beta-2)...(\beta+-\gamma+1)(\beta-\gamma)+1=\delta!$ .
When $\alpha!,\beta!,\gamma!$ are equal. $\gamma!(\alpha)(\alpha-1)(\alpha-2)...(\alpha-\gamma+1)(\alpha-\gamma)+(\beta)(\beta-1)(\beta-2)...(\beta+-\gamma+1)(\beta-\gamma)+1=3\gamma!$ $3\gamma!=\delta!$
Then the only different factor of $\gamma!$ and $\delta !$ is $3$ ,When $\delta! =3!, \gamma!= 2!$ the equality holds.
To prove that no other solutions when $\delta<3$ or $\delta>3$
Case 1: $\delta<3$ We know that 2!=2 since 2 can only be written as 2 positive integers that is 1+1 it will not satisfy the equation. The case for 1! is also similar as above.
Case 2: $\delta>3$ We know that $n!=n(n-1)!$ Since n>3 and $3(n-1)!< n!=n(n-1)!$ we could show that the sum of three factorials that is smaller than $\delta!$ when $\delta>3$ is impossible.

Without loss of generality let $a\leq b\leq c$ . Then we have $a!+b!+c!\leq 3c!$ . So the extra factor that $d!$ has over $c!$ is 3 or less. Note that there is no integer $k$ such that $k!=0$ . Therefore $c<d$ and so $d=2 \text{ or } 3$ and $c=1 \text{ or } 2$ .

There are only 4 possibilities left to check.

$1!+1!+1!=3$ which is not a factorial.

$1!+1!+2!=4$ which is not a factorial.

$1!+2!+2!=5$ which is not a factorial.

$2!+2!+2!=3\cdot 2!=3!$ which is a valid solution and therefore the only valid solution.

Note: this is pretty much identical to Sharky's solution but I tried to express it more naturally. The "without loss of generality" statement and the ordering of the variables are extremely useful and common tricks that reduce the complexity of the problem. They also make it easy to express the next idea that $d!$ can't be more than 3 times the largest term of the left hand side. Again, this inequality trick is another common way to reduce the scale of the solution space. Now, fortunately the solution space is so small you can use even inefficient and unintelligent ideas to find the solution. In conclusion, there is a standard bag of tricks built by experience that you should always think about using to reduce the complexity and solution space of a problem before you need to try to be clever.

What if a=b=c=d? Who says they must be different? 🙈

Solution $2!+2!+2!=3!$ is the only solution in which $a=b=c$ , as every other situation in which $a=b=c$ is such that $a!+b!+c!=3a!$ , which is only a factorial if $a=2$ .

Therefore, every other solution must involve at least two distinct positive integers among $a,b,c$ . Suppose, without loss of generality, that $a$ is the smallest value between a,b,c. Also suppose, without loss of generality, that $b\ge c$ . That being the case, there exist integers $k>0,l \ge 0$ such that $b=a+k$ and $c=a+l$ .

In that way, we have that $a!+b!+c!=a!+(a+k)!+(a+l)!$ , which is only a valid solution if $1+\frac{(a+k)!}{a!}+\frac{(a+l)!}{a!}=\frac{(a+m)!}{a!}$ for some $m>k,l$ .

Note that $1+\frac{(a+k)!}{a!}+\frac{(a+l)!}{a!} \le 2\frac{(a+k)!}{a!}+1$ . Also note that $\frac{(a+m)!}{a!} \ge (a+m)\frac{(a+k)!}{a!}$ , but since $(a+k) >1$ , we have that $(a+m) \ge 3$ , implying that it is impossible for $1+\frac{(a+k)!}{a!}+\frac{(a+l)!}{a!}=\frac{(a+m)!}{a!}$ when $m>k \ge l$ .

okay, so I literally guessed and got it right. I am an 8th grader. I am taking Algebra 1/ So, I found the answer 1...

2!+2!+2!+=3!