How many
satisfy
? We must have
and
.
Let
.
Then we must have
divisible by 12. This gives the possible values
for
,
which
implies
, or 9 mod 13. Similarly, if
,
must be divisible by 18, which leads to
, or 11 mod 19. (we
actually found 178 above by solving
and
The 3 choices mod 13 and mod 19 imply there are 9
with
.
How many
satisfy
? If
,
we must have
,
which leads to
,
or 17 mod 13.
Similarly, we get
,
or 12 mod 19. Thus we get 9
satisfying
this condition.
If we choose
at random, the chances of
getting an
that is not a witness are
.
If we do the
test 5 times, the chance of incorrectly concluding 247 is a prime is
.
[We actually did more work than necessary, ident
ifying the exact set of numbers which would lead to a wrong conclusion.
If we only want to count how many numbers there are, we could make use
of observations such as that, for any ,
an equation
will either have 3 solutions or no solutions.]