superprismatic

Your third circular number looks like 12, not 8. What is your definition of a circular number?

Yes, Fortran -- it's fast with no memory leaks. I've tried others like Python, Ruby, Haskell, and C but they all have what I consider fatal problems. For example, there is no attempt to insure that programs written and running well today will still run fine 10 years from now. When word got out that Python would be dropping its lambda calculus capability, I dropped it like a hot potato because most of my Python code used it.

In case anyone is interested, stewdeker's solution is the only solution (up to symmetries of the square). I wrote a little program to check this.

NoMoney is right on! Nice thinking! You didn't fall for the 100 red herring.

When you say "distinct", do you mean by value or form? So, should I consider 6/9, 4/6, and 2/3 as distinct?

I am a member of "People Whose Birthday Is NOT On February 29th Society", a not very exclusive club. A total of 128 people had satisfactorily proven that they were eligible to be members. I, being the society's secretary, was charged with mailing out the announcement for a recent meeting. The first 100 announcements had the correct time for the meeting (6 O'clock) but I had inadvertently put the time of 7 O'clock on the announcements sent to the remaining 28 members. At 6 O'clock on meeting day, 100 members showed up. We decided to wait for the remaining 28. Even though each of the members had proven that he was not born on February 29th, no records were kept of his actual birth day. So, to kill time, we all annnounced our birthdays and were astonished to find that, of the 100 people in attendence, we had only 80 distict birth days! My question to you is this: When the other 28 arrive and announce their birthdays, how many distinct birthdays will the entire group of 128 be expected to have? Assume that members were randomly chosen from the population of all persons whose birthday is not February 29th and that the distribution of birthdays is otherwise uniform for this population.

DeeGee is correct no matter how many wild cards there are.

The first answer I gave was done by making the assumption that the seller would always raise the low value and the buyer would always lower the high value. Since this is a ridiculous way to haggle, the opposite was probably meant in the OP.

Ooops!

Of course...

If, however, we take 'AN and BN will together contain each of the digits from 1 to 9 exactly once' to implicitly allow as many imbedded 0 digits as we like....

(3,2)(1,1)(2,4)(4,1)(4,4)(5,1)(3,4)(4,4) (1,1)(3,3)(1,1)(2,2)(2,4)(1,1)(2,3)(3,4) (4,4)(4,3)(3,4)(4,4)(1,1)(4,1)(4,4)(3,2) (5,1)(2,4)(3,2)(4,3)(2,5)(4,1)(4,3)(5,1) (4,4)(3,2)(1,1)(4,4)(3,2)(5,1)(1,3)(5,3) (3,4)(4,3)(1,3)(1,5)(5,1)(4,3)(5,3)? [/code]

I didn't get your hint but