Factorial, factorial, factorial

Algebra Level 2

$1\times2\times3\times\cdots \times6 = (1\times2\times3)\times(1\times2\times3\times4\times5)$

Extending the above to a larger number, we find that there exists an integer $N\, (>6)$ such that $1\times2\times3\times\cdots \times N = (1\times2\times3\times \cdots \times A)\times(1\times2\times3\times\cdots\times B),$ where $A$ and $B$ are both integers larger than 1.

What is the smallest $N?$

8 10 12 14

2 solutions

William Nathanael Supriadi
Jun 11, 2017

First, I make a list of the first few factorials. 1, 2, 6, 24, 120, 720, 5040, ... Next, I try every number $n$ on the choices, and calculate $n$ , $n(n-1)$ , $n(n-1)(n-2)$ , and so on, and see if I can get a number on the list above. I found that $10 \times 9 \times 8 = 6!$ . And we are done. $10! = 7! \times 6!$

same method I used!

Ali Azarnoush - 3 years, 11 months ago

Benjamin Becker
Jun 25, 2017

if $N$ itself is $p+1$ , where $p$ is a prime number, $N$ must be a factorial itself. $14-1=13$ is a prime, $12-1=11$ is a prime, $8-1=7$ is a prime. $8$ , $12$ and $14$ are no factorials. We only have to check 10. $10!$ $= 10*9*8*7!...$ $= (5*2) * (3*3) * (4*2) * 7!$ $= (3*2*5*4*3*2) * 7!$ $= 6*5*4*3*2*1 * 7!$ $= 6! * 7!$

Factorial, factorial, factorial

2 solutions

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