Neil's sum of factorials

Assume w.l.o.g. $x \leq y \leq z$ .

$\implies$ $x! + y! + z! \leq 3(z!)$

$\implies z! < w! \leq 3(z!)$ as $x, y$ and $z$ are positive integers.

$\implies w!/z! = 2$ or $3$

$\implies z! = 1$ or $z! = 2$

If $z! = 1$ then $x!$ and $y!$ also $= 1$ , giving $w! = 3$ which is impossible.

If $z! = 2$ then $x! + y! + z! = 4, 5$ or $6$ .

$\implies$ $w! = 6$ only possibility, with $x! = y! = z! = 2$

Only one quadruple is valid: $3! = 2! + 2! + 2!$ .

Consider the possible values for w:

• 1: Doesn't work because 1! cannot be the sum of 3 positive integers.
• 2: Doesn't work because 2! cannot be the sum of 3 positive integers.
• 3: Works because 3!=6 is exactly 2!+2!+2!.
• For w >= 4, x, y, z have to be at most w - 1. Since w! = w * (w-1)! and w > 3, there is no decomposition of w! to 3 smaller factorials. (Need at least w factorials, and w > 3.)

For $w \geq 4$ it is easy to see that there is no possible solution since $x,y,z <m$ and after $4!$ we cannot have lesser factorials adding up to $4!$ or above.

There is one ordered pair $w = 3$ and $x,y,z = 2$ which satisfies the given conditions.

We find that if $w<3$ then one of the variables $x,y$ or $z$ must be $0$ which does not follow the given conditions.

Therefore there is $\boxed{1}$ ordered pair satisfying the given conditions.

$w! = x! + y! + z!$ Note that, if $x = 1$ , then, since $w > 1$ , $w!$ is even, so $y! + z!$ must be odd. This can only happen if $y = 1$ or $z = 1$ but not both. WLOG, say $y = 1$ . Then, we have $w! = 2 + z!.$ But there are no ordered pairs $(w,z)$ that satisfy this Diophantine equation. Thus, $x, y, z > 1$ .

Factorials increase rapidly and, thus, any solutions will be for "small" $x, y, z$ . We see that $x = y = z = 2$ gives $w = 3$ and, above this, the orders of the LHS and RHS change and their difference increase extremely quickly.

Thus, the only such ordered pair is $(3, 2, 2, 2)$ , so the answer is $\boxed {1}$ .

No soultion for w>=4 as the process of choosing factorials doesnt add up ! so oly solution is (3,2,1,1)

The trivial solution is (w, x, y, z): (3, 2, 2, 2).

We use the case-to-case analysis.

If w = 1, 2: there are no solutions complying the given.

If w = 3, hence, we have the trivial solutions.

If w is greater than or equal to 4, since the factorial function increases more than exponentially, when we try to maximize x, y, and z as long as at least one of them must not equal to w. Doing this, the right-hand side hardly reach value of a factorial.

Hence, the answer.

Since the factorials increase too fast, it is impossible to get factorial of certain positive integer by adding factorials of smaller integers. In fact $(n+1)!$ can be written as sum of $n!$ s $(n+1)$ times. Clearly, replacing any of these terms would make the sum smaller and we would not reach $(n+1)!$ . Also this sum would still be greater than as far as we don't reduce every term (in that case we can state the whole argument using factorial of smaller integer than $n$ ). Hence, only possibility is that in a given equation, each term on R.H.S. is equal to $n!$ for some $n$ and term on the L.H.S. is $(n+1)!$ . Hence we have $(n+1)!=3*n!$ which gives $n=2$ . Thus there is only one solution to the problem: $\boxed{(3,2,2,2)}$

It is known that $w! = (w-1) \cdot (w-2) \cdot \cdot \cdot 3 \cdot 2 \cdot 1$ . Hence,

$0 \leq x \leq w;$

$0 \leq y \leq w;$

$0 \leq z \leq w;$

but $w!$ can never be $equal$ to $3 \cdot w!$ . $\Rightarrow 0 \leq x \leq (w-1);$

$0 \leq y \leq (w-1);$

$0 \leq z \leq (w-1);$

for $maximum$ of right hand side, $x = y = z = (w-1)$ .

By, solving, we can see that $w = 3$ , which means that this relation s only true for one value. Hence the answer.

Note that $w>x,y,z$ .First possibility $w=1$ , $w!=1$ which cannot be written as a sum of three positive numbers.Similarly for $w=2$ , $w!=2$ which cannot be written as a sum of three factorials.For $w=3$ , $w!=6 = 2!+2!+2!$ .For $w=4$ , $w!=24$ .As $3!+3!+3! =18$ , $4!$ cannot be written as the sum of three factorials.Similarly any factorial after $3!$ cannot be written as the sum of three factorials.Only ordered pair is $(3,2,2,2)$ .