
Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion 
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June 11th, 2016, 04:20 AM  #1 
Newbie Joined: Jun 2016 From: Europe Posts: 3 Thanks: 0  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 
June 11th, 2016, 05:00 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 896 
\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 $\displaystyle \frac{Y}{X}= \frac{X^Z}{X}= X^{n1}Z$, $\displaystyle \frac{Y}{X^2}= \frac{X^{n1}Z}{X}$, etc. until we reach $\displaystyle \frac{X^nZ}{X^n}= Z$. We then add $\displaystyle \frac{X^nZ}{X}+ \frac{X^{n1}Z}{X}+ \cdot\cdot\cdot+ Z= X^{n1}+ X^{n2}Z+ \cdot\cdot\cdot+ Z= (1+ X+ \cdot\cdot\cdot+ X^{n})Z$. 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 $\displaystyle 2\left(\frac{1 4^6}{1 4}\right)= 2\left(\frac{1 4096)}{3}\right)= 2\frac{4095}{3}= 2730$. Last edited by skipjack; June 11th, 2016 at 05:06 AM. 
June 11th, 2016, 05:12 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,270 Thanks: 1958 
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). 
June 11th, 2016, 10:31 AM  #4 
Newbie Joined: Jun 2016 From: Europe Posts: 3 Thanks: 0 
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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create, dividing, diving, formula, numbers, suming, summing 
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