Is there a positive integer n such that the fraction ( 9 n + 5 ) / ( 10 n + 3 )is not in the lowest term?I found n = 2 be a solution. Is it correct?

Question

Is there a positive integer n such that the fraction   (  9  n  +  5  )      /    (  10  n  +  3  )is not in the lowest term?I found   n  =  2 be a solution. Is it correct?

tennispopj8 · Accepted Answer

Your example   n  =  2 is correct. We will find all possible examples.Suppose that   d divides   9  n  +  5 and   10  n  +  3. Then   d divides  10  (  9  n  +  5  )  −  9  (  10  n  +  3  )  ,so   d divides   23Since   23 is prime, the only conceivable (positive) common divisors of   9  n  +  5 and   10  n  +  3 are   1 and   23. We will find the values of   n for which   23 is a common divisor.To make sure that 23 divides both, all we need is to make sure that 23 divides   9  n  +  5. This is because from   10  (  9  n  +  5  )  −  9  (  10  n  +  3  )  =  23 we can conclude that if 23 divides   9  n  +  5. , it must divide   10  n  +  3You noticed that 23 divides   9  n  +  5 when   n  =  2. It follows that 23 divides   9  n  +  5 if and only if   n  =  2  +  23  k for some integer   k. For if 23 divides   9  n  +  5, then 23 divides   9  n  +  5  −  23, so   23 divides   9  (  n  −  2  ), and therefore   n  −  2 is a multiple of 23. Conversely, if   n  −  2 is a multiple of 23, say   n  −  2  =  23  k, then a calculation shows that 23 divides   9  n  +  5

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