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March 26th, 2016, 04:13 AM  #1 
Newbie Joined: Mar 2016 From: London Posts: 21 Thanks: 0  Integers and algebra
The integers are all the whole numbers, but including negative whole numbers and zero; therefore 7, 5, 0, 342 and 245 are all integers. In each part, find all values of n such that both expressions are integers: i) n and 10/n ii) n and (26+n)/n iii) 2n and 7/n iv) n and (n+16)/(n+1) My answers: i) 10, 5, 2, 1, 1, 2, 5 and 10 Is there a way to find them using algebra for ii), iii) and iv) ? 
March 26th, 2016, 04:54 AM  #2 
Senior Member Joined: Dec 2012 From: Hong Kong Posts: 853 Thanks: 311 Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics 
ii) Let k = (26+n)/n. Then (k1)n=26, i.e. n = 26/(k1). For n to be an integer, k1 must be a factor of 26 (or minus that). Therefore, the possible values of k1 are 1, 2, 13, 26, 1, 2, 13, 26. The corresponding values of k are 2, 3, 14, 27, 0, 1, 12, 25. iii) Let 2n=m. Then 7/n = 14/m. For 7/n to be an integer, m must be a factor of 14 (or minus that). The possible values of m are 1, 2, 7, 14, 1, 2, 7, 14. The corresponding values of n are 0.5, 1, 3.5, 7, 0.5, 1, 3.5, 7. iv) Sorry, I don't think I know how to do this systematically. I'd sketch the graph of y = (n+16)/(n+1) to narrow down the search, then use trial and error to find all the integral solutions. (I haven't taken NT before; perhaps others can do that. ) Edit: Specifically, sketch the graph and note that the function asymptotes to 1 on both the positive and negative sides. On the left side, it increases to 1. Note that when y = 0, n = 16, so any value of n smaller than 16 is ignored. Similarly, on the left side, the function decreases and asymptotes to 1 from positive infinity, so everything from the point where y = 2  i.e. n = 14  is ignored. This narrows down our search to the integers from 16 to 14. Then I'd just plug all of these into (n+16)/(n+1) to find the integer solutions. Last edited by 123qwerty; March 26th, 2016 at 05:02 AM. 
March 26th, 2016, 05:57 AM  #3  
Newbie Joined: Mar 2016 From: London Posts: 21 Thanks: 0  Quote:
iv) 16, 6, 4, 2, 0, 2, 4, 14  
March 26th, 2016, 03:32 PM  #4 
Math Team Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 
You can always do things like $\dfrac{n + 16}{n + 1} = \dfrac{(n + 1) + 15}{n + 1} = 1 + \dfrac{15}{n + 1}$. That expression is an integer iff $(n + 1)15$. 
March 26th, 2016, 05:59 PM  #5 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
In each of those fractions, the fraction will be an integer if and only if the denominator evenly divides the numerator. 1) What integers evenly divide 10? 3) What integers evenly divide 7? 2) and 4) are a little harder but you can do the division in each case: (26+ n)/n= 1+ 26/n. What integers evenly divide 26? (n+ 16)/(n+ 1)= 1+ 15/(n+ 1). What integers evenly divide 15? So what must n be? 

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