Problem 247

The three smallest prime numbers are $2$, $3$
and $5$. Their product is $2\times3\times5=30$.
Each prime number is a positive integer, and so $\dfrac{c-3}{4}$ cannot be a prime number
unless $c>3$.

The value of $\dfrac{c-3}{4}$ is an
integer exactly when $c-3$ is
divisible by $4$.

The values of $c$ for which $4\leq c\leq 20$ and $c-3$ is divisible by $4$ are $7$, $11$, $15$, and $19$.

When $c$ is equal to $7$ and $19$, the values of $\dfrac{c-3}{4}$ are $1$ and $4$, respectively. Each of these is not a
prime number.

When $c=11$, the value of expression
is $\dfrac{11-3}{4}=2$, which is a
prime number.

When $c=15$, the value of expression
is $\dfrac{15-3}{4}=3$, which is a
prime number.

Thus, the two possible values of the integer $c$ are $11$ and $15$.
Factoring the given expression, we get $21d-77=7(3d-11)$.

For all integers $d$, the value of
$3d-11$ is equal to an integer, and
so $7(3d-11)$ is the product of two
integers, $7$ and $3d-11$.

A prime number is positive and cannot be written as a product of two
integers both greater than $1$, so
$3d-11=1$.

If $3d-11=1$, then $3d=12$, and so $d=4$.

The only integer $d$ with $1\leq d \leq10$ for which $21d-77$ is a prime number is $d=4$, and the prime number is $21(4)-77=7$.

We can confirm that if $d<4$,
$3d-11$ is negative and so $7(3d-11)$ is not prime. Further, if $d>4$, $3d-11\geq3(5)-11=4$ and so $7(3d-11)$ is the product of two integers,
both greater than $1$, and thus is
also not prime.