/sci/ - Science & Math

[Anonymous No.10615427]

How do I do this? Does anyone have a hint?

[Anonymous No.10615429]

>>10615427
Some anon already provided a very neat solution. Note that gcd(n, m) = an + bm for some integers a and b. The rest is trivial.

[Anonymous No.10615435]

and your second hint is that [math] \frac{m}{n} {n\choose m} = {n-1\choose m-1} [/math]

[Anonymous No.10615469]

>>10615427
We know that n will always be divisible by gcd(m,n), thus we know that the term gcd(m,n)/n will be less or equal to 1 and the inverse of an integer divisible by n. Futhermore, we know that n choose n is divisible by n, thus we can multiply that by an inverse of a number already divisible by n to always get an integer greater or equal to 1.

[Anonymous No.10615500]

>>10615427
What does this mean?

[Anonymous No.10615507]

>>10615500
Binomial Coefficient

[Anonymous No.10615513]

>>10615500
Choose fuction/combination
Read as n choose m :=
n!/ ((n-m)! * m!)

[Anonymous No.10615516]

>>10615469
>we know that n choose m is divisible by n
No we don't. Take n = 4, m = 2

[Anonymous No.10615570]

>>10615429
wow thanks, I don't know why I didn't see that. It's pretty easy from there yeah.

[Anonymous No.10615611]

>>10615427
why 2hu is in /sci/?