[superprismatic]

Is [(2ⁿ-1)(2ⁿ-2)(2ⁿ-2²)(2ⁿ-2³)···(2ⁿ-2^n-1)] ÷ n! an integer? Why or why not?

[bonanova]

It is.

Certainly for small values of n.

[Guest]

An Integer.

I think

for all values of n

n! is always small compared to numerator

so always we shud get an integer value...

[Guest]

The product is (1/n!)Product[2^n - 2^k,{k,0,n-1}] = Product[(2^n - 2^k)/(k+1),{k,0,n-1}]

[Guest]

Is [(2ⁿ-1)(2ⁿ-2)(2ⁿ-2²)(2ⁿ-2³)···(2ⁿ-2^n-1)] ÷ n! an integer? Why or why not?

My process

Reversed order of numerator and factored out 2's to get:

2^(n-1)*(2-1)*2^(n-2)*(2²-1)....2²*(2^(n-2)-1)*2*(2^(n-1)-1)*(2ⁿ-1).

=2^{(n^2-n)}*1*3*7*15*31*63*127*255*511*1023*2047...(2ⁿ-1)

Factoring out the 2's from the denominator, we will have a maximum of 2^(n-1) so we only need worry about the odd factors in n!:

n! with 2's factored out =1*3*5*3*7*9*5*11 ....

If n=11, then 11 must be a factor of the numerator for the results to be an integer but this is not the case so this does not always result in an integer.

[Guest]

No idea why, or more likely i've misunderstood the problem, but converting the function into excel gives me some decimals pretty quickly

[Guest]

1023 is evenly divisible by 11.

i suspect its the case that for every prime number p;

2^(p-1)-1 is evenly divisible by p.

can anyone prove this?

[superprismatic]

1023 is evenly divisible by 11.

i suspect its the case that for every prime number p;

2^(p-1)-1 is evenly divisible by p.

can anyone prove this?

That is known as Fermat's Little Theorem. You can find a simple proof on line here.

[Guest]

1023 is evenly divisible by 11.

i suspect its the case that for every prime number p;

2^(p-1)-1 is evenly divisible by p.

can anyone prove this?

Thank you for correcting me. When I first started writing my response, I meant it as only a beginning then thought I saw one that didn't work. I acted too quickly without checking every possibility. I will try to be more careful next time.