Question

For how many integers N between 1 and 1990 is the improper fraction (N2 + 7) / N + 4 not in the lowest terms? (A) 0 (B) 86 (C) 90 (D) 104

Answer with proper explanation. ​

Answer 1

Answer:

Firstly, for the fraction not to be in it's lowest terms both the numerator and denominator must be divisible by some positive integer ,say k

Hence we must have that:

gcd (N²+7,N+4)≠1

Thus

gcd (N²+7,N+4)=> gcd (4N-7,N+4)

=>gcd (23,N+4)= 1 or 23 (since 23 is a prime)

We discard 1 since it would give the fraction in its lowest terms

Hence for gcd(N²+7,N+4)=>gcd(23,N+4)=23

We must have that both N²+7 and N+4 are divisible by 23

Thus , we look for the number of Integer multiple of 23 less than 1990

23m <1990 ( where m=1,2,3---------)

We see that m=86

So there exist 86 integers between 1 and 1990 for this to hold

Answer 2

Given:

$(N^2 + 7) / N + 4$

To Find:

how many integers N between $1$ and $1990$.

Step-by-step explanation:

Since $n^2+7=(n+4)(n-4)+23$

GCD $(n+4,n^2+7)=$gcd$(n+4,23)$∈$1,23$.

Thus$\frac{n^2+7}{n+4}$ is not the lowest form

If and only if$23|(n+4).$.So $n$ is of the form $23k-4$ and $1\leq n\leq 1990$ translates to $1\leq k\leq \leq \leq 86$. So there are$86$ such integers given by $n$∈$23k-4:1\leq k\leq 86$.

Answer:

Thus the $1$ and $1990$ improper fraction is $86$.