Unique representation: $a!cdot b!cdot c!=m!cdot n!cdot p!$

Mathematics Asked by felipeuni on December 6, 2020

Let $$a,b,cge 3$$ be natural numbers. If $$ale ble c$$ and $$mle nle p$$ the following proposition is true?

$$a!cdot b!cdot c!=m!cdot n!cdot p!to (a,b,c)=(m,n,p)$$

$$a!cdot b!=m!cdot n!to (a,b)=(m,n)$$

Any hints would be appreciated.

Combine $$6!7!=10!$$ with @N.S.'s answer

$$6!7!(n!)!=10!n!(n!-1)!$$

• $$6!7!24!=4!10!23!$$
• $$6!7!120!=5!10!119!$$
• $$6!7!720!=6!10!719!$$

Answered by cosmo5 on December 6, 2020

The famous $$n! cdot (n!-1)!=(n!)!=(n!)! cdot 1!$$ is a counterexample for 2, and can be extended to $$3$$:

$$1!cdot1!cdot((n!)!)!=(n!)! ((n!)!-1)!=n! cdot(n!-1)! cdot((n!)!-1)!$$

Answered by N. S. on December 6, 2020

$$4!4!25! = 5!5!24!$$

$$5!5!36! = 6!6!35!$$

$$6!4!35! = 7!5!34!$$ ....

$$a!b![(a+1)(b+1)]! = (a+1)!(b+1)![(a+1)(b+1)-1]!$$

Answered by Mike on December 6, 2020

The part about two is wrong

$$4! 15! = 7! 13!$$

$$3! 20! = 5! 19!$$

$$4! 30! = 6! 29!$$

$$5! 42! = 7! 41!$$

$$6! 56! = 8! 55!$$

$$18! 57! = 22! 54!$$

$$7! 66! = 14! 62!$$

$$7! 72! = 9! 71!$$

$$8! 90! = 10! 89!$$

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$$3! 3! 16! = 4! 4! 15! = 4! 7! 13! = 2^{17} cdot 3^8 cdot 5^3 cdot 7^2 cdot 11 cdot 13$$ $$3! 8! 13! = 4! 6! 14! = 2^{18} cdot 3^8 cdot 5^3 cdot 7^2 cdot 11 cdot 13$$

$$3! 4! 20! = 4! 5! 19! = 2^{18} cdot 3^8 cdot 5^3 cdot 7^2 cdot 11 cdot 13$$

Answered by Will Jagy on December 6, 2020