Find 9 is really a bad name for this one but it is similar in that we are looking for numbers whose last digit divided into itself results in 0.

Using the set of digits 1-9 and input of the number of digits (always starting with 1 and going in sequence) and the grouping of digits, create a method that will return a count of all the permuations such that each sequetial group meets the rule stated above. For example:

637284591 (group by 3)

637/7 has no remainder.

372/2 has no remainder.

728/8 has no remainder.

284/8 has no remainder.

845/5 has no remainder.

459/9 has no remainder.

591/1 has no remainder.

The correct count if digits were 3 and group were 2 would be 2 as follows:

123 false;

132 false;

312 true; (31/1 and 12/2)

321 true; (32/2 and 21/1)

231 false;

213 false;

As a

test (and I'm not 100% sure about this

) for 6 digits grouped by 3 I got 136 permutations that met the criteria.

This one is worth a box of Peanut Butter cookies Eric. You'll have to use a form for this one.

[ June 01, 2003: Message edited by: Michael Morris ]