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

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The discussion focuses on finding all triples (a, b, c) that satisfy the equation a! = 4(b!) + 10(c!). Several solutions are provided, including (4, 1, 2), (6, 5, 4), and (14, 13, 3). Key observations include that a must be greater than both b and c, and assuming c = b leads to a specific solution. The equation can be manipulated to explore different relationships between a, b, and c, guiding readers on how to approach the problem. The conversation encourages further exploration and completion of the solution process.
Marcelo Arevalo
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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.
 
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Hi Marcelo Areyalo!

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

Marcelo Arevalo said:
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
 
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