A number theory problem by Matin Naseri

Number Theory Level 2

$\text{A!}={A!!}$

What is the smallest even possible value for the $\text{(A)}$ such that $\text{(A>0)}$ ?

1 solution

Matin Naseri
Feb 2, 2018

According to above $\text{Wiki}$ we get that.

$\text{2!=2×1=2}$

$\text{2!!=2}$ .

Thus $\text{A}$ equal to $\large{2}$

For $A = 1$

$1! = 1!!$

where $1 < 2$

I think you mean ''smallest even possible positive value of $A$ ...''

Munem Shahriar - 3 years, 4 months ago

Oh sorry i will fix it.

Matin Naseri - 3 years, 4 months ago

Mr. shahriar!

smallest even possible is $\text{0}$ .

$\text{(0!=1)}={(0!!=1)}$

Just for known .

Matin Naseri - 3 years, 4 months ago

@Matin Naseri – The problem says ''Hint: $A$ is a positive integer''. But 0 is not a positive integer.

Munem Shahriar - 3 years, 4 months ago

@Munem Shahriar – I have deleted the $\text{Hint}$ and replace with $\text{A > 0}$ .

Matin Naseri - 3 years, 4 months ago