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March 26th, 2016, 04:13 AM   #1
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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) ?
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March 26th, 2016, 04:54 AM   #2
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ii) Let k = (26+n)/n. Then (k-1)n=26, i.e. n = 26/(k-1). For n to be an integer, k-1 must be a factor of 26 (or minus that). Therefore, the possible values of k-1 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.
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Last edited by 123qwerty; March 26th, 2016 at 05:02 AM.
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March 26th, 2016, 05:57 AM   #3
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Quote:
Originally Posted by 123qwerty View Post
ii) Let k = (26+n)/n. Then (k-1)n=26, i.e. n = 26/(k-1). For n to be an integer, k-1 must be a factor of 26 (or minus that). Therefore, the possible values of k-1 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.
Thanks, i did it how you explained and I got:

iv) -16, -6, -4, -2, 0, 2, 4, 14
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March 26th, 2016, 03:32 PM   #4
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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 $(n + 1)|15$.
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March 26th, 2016, 05:59 PM   #5
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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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