Creating Even Problems with Restrictions: x!

[Elihu5991]

I have been given a problem to make a problem for x!. The only restriction is, "But omit odd numbers given x is even"

Another is for x! is, "Omit even numbers, x is divisible by 3".

Is it even possible? I've thought of several manners, but I don't think I'm correct.

Thanks in advance, everyone.

 

[Simon Bridge]

What do you mean "make a problem"? Are you supposed to design a math question?
If I don't know your intended result I cannot help you. How would you know if you are "correct"?

note, if n is a positive integer, then x_n=2n-1 is an odd number.
Does that help?

 

[pwsnafu]

I have been given a problem to make a problem for x!.

What do you mean "make a problem"? Is that a typo?

I think what you are asking for is a formula for, say,
$$1 \cdot 3 \cdot 5 \cdot 7 \cdot 9$$
Okay, start with 9!
How do you construct the product above from 9 factorial?

 

[HallsofIvy]

those are sometimes denoted (x)!. If x is even it is x(x- 2)(x- 4)...(4)(2). If x is odd, it is x(x- 2)(x- 4)...(3)(1).

Note that is x is even, say x= 2n, then we have (2n)(2n-2)(2n- 4)... (4)(2)= (2)(n)(2)(n-1)(2)(n- 2)...(2)(2)(2)(1)= n(n-1)(n-2)...(2)(1) times a power of 2.

 

[Elihu5991]

Thankyou so much everyone! This problem was a while back. I have had problems accessing this forum, that's why I haven't replied in a long time.

If I can remember what was the problem (I don't normally forget things), as it may have been a typo, then I can reiterate the issue. I still want it solved, but till then, thanks.

 

[Simon Bridge]

Another is for x! is, "Omit even numbers, x is divisible by 3".

omit even means only odd numbers: 1,3,5,7,9,11,13,15,17
but only divisible by three means that 1,5,7,11,13,17... are not allowed; which leaves: 3, 9, 15, ... which is 3(1,3,5,...) ...

i.e. it is another way of asking for the product of odd-integer multiples of three.
So-

If x is an odd-integer multiple of 3, then the product of odd-integer multiples of 3 up to x is given by:$$3\prod_{n=1}^{x/3}(2n-1) = 3\left ( \frac{x}{3} \right )!$$... isn't it?

This sort of word problem is an exercise in curly thinking.