# I have found a formula for dividing numbers in easy steps

I found an easy method for division and it depends on some factors.

I wanted to find an answer for $1000/101$ with easy steps. My starting point is here. I formulated this method by 2 hours of hard work. It is an infinite series, but taking 4 or 6 units of the series we can arrive at an answer easily.

# To find $100/11$

We know $100/10 = 10$, then $100/11 = 10 - 1 + 0.1 - 0.01$ (each term is the preceding term divided by $10$; division by $10$ is easy) $= 9 + 0.09$ (simplify by taking two members of the series, so easy steps) $= 9.09$

# To find $100/12$

We know $100/11 = 9.09$, then $100/12 = 9.09 - 0.909 + 0.0909 - 0.00909$ (each member of the series is found by dividing by $10$ with the preceding member) giving $8.181+.08181$ (calculations are easy) or $8.29$

I took only $4$ members of the series, and if we take $6$ members of the series, we get a better result.

# To find $1000/101$

We know $1000/100 = 10$, then $1000/101 = 10 - 0.1 + 0.001 - .0001 = 9.9 + 0.0009 = 9.9009$

# To find $1000/102$

We know, $1000/101 = 9.9$ approximately, then $1000/102 = 9.9 - 0.099 + 0.00099- 0.0000099 = 9.801 + 0.0009801 = 9.801$ (approx)

# To find $100/3$

We know $100/2 = 50$, then $100/3 = 50 - 25 + 12.5 - 6.25 + 3.125 - 1.5625 = 25 + 6.25 + 1.5625 = 32.8125$ or approximately $33$.

Since they are a small number to get a perfect answer (we need to take more members of the series; for big numbers $4$ members of the series is sufficient)

Has somebody found this before me? Where should I submit the infinite series that I have found for further evaluation?

• This is a good one; it should go into the next edition of Dead Reckoning. You may want to get a copy of that book; it is full of such approximations. – richard1941 Sep 12 '18 at 5:49

What you are using are so called geometric series: $$\frac1{1+n}=\frac1n\frac1{1-(-1/n)}=\frac1n\sum_{k=0}^{\infty}\left(-\frac1n\right)^k$$ thus $$\frac{x}{1+n}=\frac xn\sum_{k=0}^{\infty}\left(-\frac1n\right)^k=\frac{x}n-\frac{x}{n^2}+\frac{x}{n^3}-\dots.$$