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June 11th, 2016, 04:20 AM   #1
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Trying to create formula for dividing and summing numbers

Hello all, I am trying to create a formula considering having following variables.
I would appreciate if you could help me understand this.

For an example: I have a number 243 and I have a dividing factor of 3 and I have end point factor which is 1 in this case.

I would need:

243 3
81 3
27 3
9 3
3 3
1 3


And in the end sum all numbers n1+n2+n3+n4, which in this case is 364


or if I have a number 2048, diving factor of 4 and end point 2.

2048 4
512 4
128 4
32 4
8 4
2 4

At the end sum result is 2730.

Like I said: I have a number, dividing factor and end number. Now I would really like if you could explain me how to get a formula for this. I would really appreciate it.

Last edited by skipjack; June 11th, 2016 at 05:09 AM. Reason: diving -> dividing
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June 11th, 2016, 05:00 AM   #2
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\First, I am pretty sure you mean "dividing" rather than "diving". So have a dividend Y and a divisor X and you divide X into Y repeatedly until it will no longer divide evenly. We can factor $\displaystyle Y= X^nZ$ where Z includes all factors of Y other than X.

Each quotient is $\displaystyle \frac{Y}{X}= \frac{X^Z}{X}= X^{n-1}Z$, $\displaystyle \frac{Y}{X^2}= \frac{X^{n-1}Z}{X}$, etc. until we reach $\displaystyle \frac{X^nZ}{X^n}= Z$.
We then add $\displaystyle \frac{X^nZ}{X}+ \frac{X^{n-1}Z}{X}+ \cdot\cdot\cdot+ Z= X^{n-1}+ X^{n-2}Z+ \cdot\cdot\cdot+ Z= (1+ X+ \cdot\cdot\cdot+ X^{n})Z$.

That sum is a "finite geometric sum" which can be shown to be equal to $\displaystyle \frac{1- X^{n+1}}{1- X}$.

So your formula is $\displaystyle Z\left(\frac{1- X^{n+1}}{1- X}\right)$.

In the case of Y= 243, X= 3, $\displaystyle Y= 3^5$ so n= 5 and Z= 1. The formula above give $\displaystyle 1\left(\frac{1- 3^6}{1- 3}\right)= \frac{1- 729}{1- 3}= \frac{-728}{-2}= 364$.

Similarly, if Y= 2048 and X= 4, $\displaystyle Y= 4^5(2)$ so n= 5 and Z= 2. The formula above gives $\displaystyle 2\left(\frac{1- 4^6}{1- 4}\right)= 2\left(\frac{1- 4096)}{-3}\right)= 2\frac{4095}{3}= 2730$.
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Last edited by skipjack; June 11th, 2016 at 05:06 AM.
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June 11th, 2016, 05:12 AM   #3
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If the factor is r and the first and last numbers are a and b,
multiply the sum by (r - 1) and notice how most terms cancel.

It follows that the formula for the sum is (ar - b)/(r - 1).
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June 11th, 2016, 10:31 AM   #4
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Countryboy and SkipJack.

Thank you very much, I have tested both of your formulas. They work excelent. Great minds. Thank you.
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