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, identifying 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.]

2002-12-14