Let there be N notepads. Each divisors of N unequal to 1 or N gives
one reasonable way to pile the notebooks. Hence N has 3 such divisors, or
5 divisors in all. It follows that N = p^4 for some prime p.
Since 10000 < N < 500000, p must be 11, 13, 17, 19, or 23. Marcia's slip
of paper had p, p^2, p^3, and p^4 written on it. Let's tabulate these values
for each possible value of p:

p p^2 p^3 p^4
11 121 1331 14641
13 169 2197 28561
17 289 4913 83521
19 361 6859 130321
23 529 12167 279841

The slip of paper had a zero on it, so p must be 19 and there were
130321 notepads.

David Moews
(
dmoews@fastmail.fm )

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