Theorem (1): The product of any n consecutive integers is divisible by 
.
Proof (1):
For 
, and to show that the last expression is an integer it is sufficient to show that any prime p which occurs in 
to at least as high as a power in
. Thus, we have to show that
![I[\frac{(m+n)}{p}]+I[\frac{(m+n)}{p^{2}}]+I[\frac{(m+n)}{p^{3}}]+\ldots](https://s0.wp.com/latex.php?latex=I%5B%5Cfrac%7B%28m%2Bn%29%7D%7Bp%7D%5D%2BI%5B%5Cfrac%7B%28m%2Bn%29%7D%7Bp%5E%7B2%7D%7D%5D%2BI%5B%5Cfrac%7B%28m%2Bn%29%7D%7Bp%5E%7B3%7D%7D%5D%2B%5Cldots+&bg=ffffff&fg=000&s=0 "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](https://s0.wp.com/latex.php?latex=I%5Bm%2Fp%5D%2BI%5Bm%2Fp%5E%7B2%7D%5D%2BI%5Bm%2Fp%5E%7B3%7D%5D%2B%5Cldots+%2BI%5Bn%2Fp%5D%2BI%5Bn%2Fp%5E%7B2%7D%5D%2BI%5Bn%2Fp%5E%7B3%7D%5D%2B%5Cldots&bg=ffffff&fg=000&s=0 "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]](https://s0.wp.com/latex.php?latex=I%5Bx%2Fy%5D&bg=ffffff&fg=000&s=0 "I[x/y]")
: If a is a fraction or an irrational number, the symbol 
will be used to denote the integral part of a.
Now, ![I[\frac{(m+n)}{p}] \geq I[m/p]+I[n/p]](https://s0.wp.com/latex.php?latex=I%5B%5Cfrac%7B%28m%2Bn%29%7D%7Bp%7D%5D+%5Cgeq+I%5Bm%2Fp%5D%2BI%5Bn%2Fp%5D&bg=ffffff&fg=000&s=0 "I[\frac{(m+n)}{p}] \geq I[m/p]+I[n/p]")
, and the same is true if we replace p by 
, 
in succession, hence the result in question.
Theorem (2): If n is a prime, then 
is divisible by n.
For by the preceding 
is divisible by 
and since n is a prime and r is supposed to be less than n, is prime to n. Hence,
is a divisor of 
and 
is divisible by n.
Thus, if a is a prime, all the coefficients in the expansion of 
except the first and last are divisible by n.
Homework:
1) Find the highest power of 5 contained in 
2) If n is an odd prime, the integral part of 
is divisible by 
.
More later,
Nalin Pithwa
Like this:
Like Loading...
Related
