How Can I Find All Triples (a, b, c) in This Equation?

[Marcelo Arevalo]

I came across this problem in a book that I am reding.

Find All Triples (a, b, c) if a! = 4(b!) + 10(c!)there is an answer key :
(4, 1, 2) (6, 5, 4) (14, 13, 3)sorry I am not being lazy to solve this.. the truth is I don't know how to begin with in solving this kind of problem.
can you please help me or guide me on how to do this??
thank you everyone.

 

[anemone]

Hi Marcelo Areyalo!

Here are a few observations that might help to get you started:

I came across this problem in a book that I am reding.

Find All Triples (a, b, c) if a! = 4(b!) + 10(c!)

there is an answer key :
(4, 1, 2) (6, 5, 4) (14, 13, 3)

1.

$a$ is certainly greater than both $b$ and $c$.

2.

Assume that $c=b$, we then get
$\begin{align*}a!&= 4(b!) + 10(c!)\\&=14b!=14(13!)=14!\implies (a,\,b,\,c)=(14,\,13,\,13)\end{align*}$

3.

Let $c\gt b$, the equality can be rewritten as:

$a!= 4(b!) + 10(c!)$

$\small 1(2)(3)\cdots(b-1)(b)\cdots(c-1)(c)\cdots(a-1)(a)= 4(1)(2)(3)\cdots(b-1)(b) + 10(1)(2)(3)\cdots(b-1)(b)(b+1)\cdots(c-1)(c)$

$\small \dfrac{1(2)(3)\cdots(b-1)(b)(b+1)\cdots(c-1)(c)\cdots(a-1)(a)}{(1)(2)(3)\cdots(b-1)(b) }=\dfrac{ 4(1)(2)(3)\cdots(b-1)(b)}{(1)(2)(3)\cdots(b-1)(b) } + \dfrac{10(1)(2)(3)\cdots(b-1)(b)(b+1)\cdots(c-1)(c)}{(1)(2)(3)\cdots(b-1)(b) }$

$\therefore (b+1)\cdots(c-1)(c)\cdots(a-1)(a)=4+ 10(b+1)\cdots(c-1)(c)$

I will leave it to you and the readers to complete the rest. :D