Some basic results in number theory — part 1

Theorem (1): The product of any n consecutive integers is divisible by n!.

Proof (1): 

For \frac {(m+1)(m+2) \ldots (m+n)}{n!}=\frac{(m+n)!}{m!n!}, and to show that the last expression is an integer it is sufficient to show that any prime p which occurs in m!n! to at least as high as a power in

(m+n)!. Thus, we have to show that

I[\frac{(m+n)}{p}]+I[\frac{(m+n)}{p^{2}}]+I[\frac{(m+n)}{p^{3}}]+\ldots is greater than or equal to

I[m/p]+I[m/p^{2}]+I[m/p^{3}]+\ldots +I[n/p]+I[n/p^{2}]+I[n/p^{3}]+\ldots.

Note: The  Symbol I[x/y]: If  a is a fraction or an irrational number, the symbol I(a) will be used to denote the integral part of a.

Now, I[\frac{(m+n)}{p}] \geq I[m/p]+I[n/p], and the same is true if we replace p by p^{2}, p^{3} in succession, hence the result in question.

Theorem (2): If n is a prime, then {n \choose r} is divisible by n.

For by the preceding n(n-1)(n-2)\ldots (n-r+1) is divisible by r! and since n is a prime and r is supposed to be less than n, r! is prime to n. Hence,

r! is a divisor of (n-1)(n-2)\ldots (n-r+1)

and n(n-1)\ldots (n-r+1)/r! is divisible by n.

Thus, if a is a prime, all the coefficients in the expansion of (1+x)^{n} except the first and last are divisible by n.


1) Find the highest power of 5  contained in 158!

2) If n is an odd prime, the integral part of (\sqrt{5}+2)^{n}-2^{n+1} is divisible by 20n.

More later,

Nalin Pithwa

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: