1. The problem asks to find the greatest common divisor (GCD) for two pairs of integers involving the variable $n$. 2. For part (a), find $\gcd(-7, \ 2n + 3)$. - Since $-7$ is a constant and prime, the GCD must be a divisor of 7. - The possible divisors are $1, 7, -1, -7$, but GCD is always taken as a positive integer. - Thus, either $\gcd = 1$ or $7$. - $7$ divides $2n + 3$ exactly means $2n + 3 \equiv 0 \pmod{7}$. 3. Solve the congruence for $n$: $$ 2n + 3 \equiv 0 \pmod{7} \implies 2n \equiv -3 \equiv 4 \pmod{7} $$ Since $2 \times 4 = 8 \equiv 1 \pmod{7}$, the inverse of 2 modulo 7 is 4. Multiply both sides by 4: $$ n \equiv 4 \times 4 = 16 \equiv 2 \pmod{7} $$ So $n$ must satisfy $n \equiv 2 \pmod{7}$ for the GCD to be 7. 4. For part (b), find $\gcd(9, \ 4 - n)$. - $9 = 3^2$, so possible divisors are $1, 3, 9$. - To have $\gcd = 9$, $9$ must divide $(4 - n)$ exactly: $$ 4 - n \equiv 0 \pmod{9} \implies n \equiv 4 \pmod{9} $$ - To have $\gcd = 3$, $3$ must divide $(4 - n)$ but $9$ does not: $$ 4 - n \equiv 0 \pmod{3} \implies n \equiv 1 \pmod{3} $$ but $n \not\equiv 4 \pmod{9}$. - Otherwise, $\gcd = 1$. Final answers: - Part (a): - $\gcd = 7$ if $n \equiv 2 \pmod{7}$ - Else, $\gcd = 1$ - Part (b): - $\gcd = 9$ if $n \equiv 4 \pmod{9}$ - $\gcd = 3$ if $n \equiv 1 \pmod{3}$ and $n \not\equiv 4 \pmod{9}$ - Else, $\gcd = 1$