# How to prove that $a \cdot b$ is not divisible by 5 for $\frac{1}{1} + \frac{1}{2} + … + \frac{1}{99} + \frac{1}{100} = \frac{a}{b}$? [duplicate]

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.

## marked as duplicate by vrugtehagel, Aretino, joriki, Matthew Conroy, ShaileshMar 11 '16 at 23:32

• Well, $\frac11+\frac12+\cdots+\frac1{100}=\frac{14466636279520351160221518043104131447711}{2788815009188499086581352357412492142272}$ so yeah, $a\cdot b$ is not divisible by $5$. – vrugtehagel Mar 11 '16 at 21:03
• without a computer :) – piteer Mar 11 '16 at 21:04
• Is $a.b$ the same as $a\times b$ – Shuri2060 Mar 11 '16 at 21:06

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.