How to prove that $a \cdot b$ is not divisible by 5 for $\frac{1}{1} + \frac{1}{2} + ... + \frac{1}{99} + \frac{1}{100} = \frac{a}{b}$? 
Let  $$\frac{1}{1}  +  \frac{1}{2}  + ... + \frac{1}{99}  + \frac{1}{100}  =  \frac{a}{b},$$  where $a,b$ natural numbers and  $\gcd(a,b) = 1$. How to prove that $a \times b$ is not divisible by $5$?

I know $\frac{1}{5k+1}+\frac{1}{5k+2}+\frac{1}{5k+3}+\frac{1}{5k+4}=\frac{a}{b}$,$k \in N$ ,where a ,b relatively prime,so a is  divisible by 25, and b is not  divisible by 5.
 A: If you know $\frac{1}{5k+1}+\frac{1}{5k+2}+\frac{1}{5k+3}+\frac{1}{5k+4}=\frac{a}{b}$ with $25|a$.  You pretty much have it.
Let  $\frac{1}{5k+1}+\frac{1}{5k+2}+\frac{1}{5k+3}+\frac{1}{5k+4}=25\frac{a_k}{b_k}$ where $5 \not \mid a_k$ and $5\not \mid b_k$.
Consider $1/1 + ... + 1/4 = 25\frac{a_0}{b_0}$ and $1/6+ .... + 1/9 = 25\frac{a_1}{b_1}$ up to $1/96 + .. + 1/99 = 25\frac{a_{19}}{b_{19}}$. You can add these  to get 25$\sum \frac{a_i}{b_i}= 25\frac A B$. As $a_i$ and $b_i$ are coprime to 5, A and B are co prime to 5.
Now Consider $1/5 + 1/10 + .. 1/20 = 1/5(1 + 1/2 + 1/3 + 1/4)$ up to $1/80 + .... + 1/95 = 1/5(1/16 + ... + 1/19)$.  These each add up to $5*\frac{a_i}{b_i}$ (for $i=0..3$) As above we can add all these terms to $5\frac C D$ where $C$ and $D$ are coprime to 5. 
Finally Consider $1/25 + 1/50 + 1/75 + 1/100 = 1/25(1 + 1/2 + 1/3 + 1/4) = \frac {a_0}{b_0}$.  Let's rewrite the variables as $E = a_0$ and $F = b_0$.
So $1/1 + ... + 1/100$ = $25A/B + 5C/D + E/F$ where 5 doesn't divide any of the variables:
$\dfrac{25A\gcd(BDF)/B + 5C\gcd(BDF)/D + E\gcd(BDF)/F}{\gcd(BDF)}$.  5 does not divide the denominator.  But as 5 divides 25A and 5C but not E, 5 doesn't divide the numerator either.
