Is there a positive integer n such that the fraction ( 9 n + 5 ) <mrow class="MJX-TeX

Roland Manning

Roland Manning

Answered question

2022-06-23

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?

Answer & Explanation

tennispopj8

tennispopj8

Beginner2022-06-24Added 20 answers

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 23
Since 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 + 3
You 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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