Three prime numbers $p$, $q$ and $r$, each less than 20, are such that $p–q=q–r$. How many distinct possible values can we get for $(p+q+r)$?

The correct option is 4.

Let three distinct or non-distinct prime numbers be $p$, $q$, and $r$.

We are given that $p$, $q$, and $r$ are all prime numbers less than 20.

The prime numbers less than 20 are:

$2,3,5,7,11,13,17,19$

We are given the condition:

$p-q=q-r$

Rearranging the terms, we get:

$p+r=2q$

Thus, the sum $p+q+r$ can be simplified as:

$p+q+r=(p+r)+q=2q+q=3q$

Since , the numbers $r$, $q$, and $p$ form an arithmetic progression with a common difference $d=q-r=p-q$.

Without loss of generality, assuming $r<q<p$ (so $d>0$), let us test each prime number $q$ from our list to see if there exist prime numbers $r$ and $p$ less than 20 such that $q-r=p-q$:

1. If $q=3$:

The only prime less than 3 is 2 ($r=2$), giving $d=1$. Then $p=3+1=4$, which is not prime.

2. If $q=5$:

Possible primes $r<5$ are 2 and 3.

If $r=3$, then $d=2$, so $p=5+2=7$ (prime). Here, $(r,q,p)=(3,5,7)$.

Value of $p+q+r=3\times 5=15$.

3. If $q=7$:

If $r=3$, then $d=4$, so $p=7+4=11$ (prime). Here, $(r,q,p)=(3,7,11)$.

Value of $p+q+r=3\times 7=21$.

4. If $q=11$:

If $r=3$, then $d=8$, so $p=11+8=19$ (prime). Here, $(r,q,p)=(3,11,19)$.

If $r=5$, then $d=6$, so $p=11+6=17$ (prime). Here, $(r,q,p)=(5,11,17)$.

Value of $p+q+r=3\times 11=33$.

5. If $q=13$:

If $r=7$, then $d=6$, so $p=13+6=19$ (prime). Here, $(r,q,p)=(7,13,19)$.

Value of $p+q+r=3\times 13=39$.

6. If $q=17$ or $q=19$:

For any prime $r<q$, the value of $p=2q-r$ will be greater than or equal to $2\left(17\right)-13=21$, which is greater than 20.

Thus, the possible values for $q$ are $5,7,11,$ and $13$.

The distinct possible values for $(p+q+r)=3q$ are:

$15,21,33,\text{ and }39$

Therefore, there are 4 distinct possible values for $(p+q+r)$.