
Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion 
 LinkBack  Thread Tools  Display Modes 
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: 894 
\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: 19,963 Thanks: 1849 
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. 

Tags 
create, dividing, diving, formula, numbers, suming, summing 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Please help to create formula on use electricity by clients and solar system  valet  Advanced Statistics  0  October 28th, 2014 03:02 PM 
Deriving the formula for summing a sequence of squares  Mr Davis 97  PreCalculus  11  August 17th, 2014 10:05 AM 
Dividing Very Large Numbers  nogar  Elementary Math  1  March 14th, 2014 04:39 PM 
Help to create ranking formula!  demec  Real Analysis  2  November 29th, 2009 11:59 AM 
Dividing 10^100,000 with 10^1000 numbers  godilya  Number Theory  5  September 5th, 2009 01:50 PM 