, 2024

We begin by recognizing that $2$ is the only even prime number.

If $x$, $y$ and $z$ are each odd prime numbers, then both
$x+y$ and $x+z$ are even prime numbers (since the sum
of two odd numbers is even).

However, both $x+y$ and $x+z$ are each at least $2+3=5$, and therefore each must be an odd
prime number.

This tells us that $x,y,z$ cannot all
be odd prime numbers, and so exactly one of them is equal to $2$ (since they are all different from one
another and $2$ is the only even
prime number).

If $y=2$, then each of $x$ and $z$ is odd and so $x+z$ is even, which is not possible.

Similarly, if $z=2$, then each of
$x$ and $y$ is odd and so $x+y$ is even, which is not possible, and
so we conclude that $x=2$.

Substituting $x=2$, the list of $8$ different prime numbers becomes: $$w, 2, y, z, 2+y, 2+z, 234+z, 234-z,$$ and
we note that $w+x+y=234$ becomes
$w+y=232$.

Since $z$ and $2+z$ are prime numbers that differ by
$2$, next we consider the consecutive
odd prime numbers with $z$ less than
$50$.

These are: $3$ and $5$, $5$
and $7$, $11$ and $13$, $17$ and $19$, $29$ and $31$, $41$ and $43$.

So then $z$ is equal to one of $3$, $5$, $11$, $17$, $29$, or $41$.

If $z=3$, then $234-z=231$ which is divisible by $3$ and thus not a prime number.

If $z=11$, then $234+z=245$ which is divisible by $5$ and thus not a prime number.

If $z=17$, then $234-z=217$ which is divisible by $7$ and thus not a prime number.

If $z=29$, then $234-z=205$ which is divisible by $5$ and thus not a prime number.

If $z=41$, then $234+z=275$ which is divisible by $5$ and thus not a prime number.

Finally, if $z=5$, then $234-z=229$ and $234+z=239$, and both of these are prime
numbers.

Alternately, we may have noted that if $z$ has units digit $1$, then $234+z$ has units digit $5$, and if $z$ has units digit $9$, then $234-z$ also has units digit $5$, and so each is divisible by $5$, which is not possible since each is a
prime number. We could have then removed $z=11,29,41$ as possibilities and
considered only $z=3,5,17$ as we did
above.

The table below summarizes what we know about the $8$ different prime numbers to this
point.

$w$
$x$
$y$
$z$
$x+y$
$x+z$
$234+z$
$234-z$

$2$

$5$
$2+y$
$7$
$239$
$229$

As shown previously, since $y$ and
$2+y$ are consecutive odd prime
numbers (with $y$ less than 50), then
$y$ is equal to one of 3, 11, 17, 29,
or 41 (recall that $z=5$ and the 8
numbers must all be different).

Since $w+y=232$, then $w=232-y$.

For which value(s) of $y$ is $w=232-y$ a prime number different from
those already in our list?

If $y=3$, then $w=229$ which is not possible since $234-z=229$.

If $y=11$, then $w=221$ which is divisible by 13 and
therefore not a prime number.

If $y=17$, then $w=215$ which is divisible by 5 and
therefore not a prime number.

If $y=29$, then $w=203$ which is divisible by 7 and
therefore not a prime number.

Finally, if $y=41$, then $w=191$ which is a prime number.

The final list of 8 different prime numbers is shown below.

$w$
$x$
$y$
$z$
$x+y$
$x+z$
$234+z$
$234-z$

$191$
$2$
$41$
$5$
$43$
$7$
$239$
$229$

The value of $w-y$ is $191-41=150$.