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Subject: "integer?"     Previous Topic | Next Topic
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Conferences The CTK Exchange High school Topic #217
Reading Topic #217
leiyou
guest
Dec-08-02, 10:13 PM (EST)
 
"integer?"
 
   Sorry, this prob may have been posted before

The problem goes like this

for what integral value of n will the expression

2*n / (n-4) be a positive integer.

obviously the n must be greater than 4

and we can show that the expression is integer for
n = 5, 6, 8, 12

however, how do we prove that no n from 13 and onwards will give the expression an integral value?


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RicBrad
Member since Nov-16-01
Dec-11-02, 01:23 PM (EST)
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1. "RE: integer?"
In response to message #0
 
   Expand your original fraction by long division:

2n / (n-4) = 2 + 8 / (n-4)

So we see that this can only be an integer if (n-4) is a factor of 8, and furthermore, it is only a positive integer if 8/(n-4) is greater than or equal to -1.

The factors of 8 are 1,2,4,8 (and -1,-2,-4,-8 for our purposes)

so n = -4,5,6,8,12

> obviously the n must be greater than 4

be careful with what is 'obvious'

Rich


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