There are 87 square numbers that use every digit (base 10, no leading zeroes)) exactly once.

List them.

(In reply to

re(3): Pencil and Paper solution (spoiler) by Jer)

Yes it does.

But you can get much closer if you extend this parameter and add another one:

a. As to the quantity of qualifying 10- digit numbers 9*9! is totally unacceptable number- the upper bond is definitely below **270451**- the number quoted in A036755 (#of 10 last distinct digits of a square -

without "no leading zeroes disclaimer") .

b. Once you divide Charlie's list into 9 sectors, considering the most significant digit,you will find how far from uniform the distribution is:

the shortest interval (starting with digit 3 claims 18 out 87 answers,- while the longest (starting with 1)- only 6.

In each of those intervals there are 9! pandigital numbers , but with different chances to be a square (comment later, below).

As to 67389 "roots" - dense in the beginning - scarcer towards the upper band.

Imagine trying to extrapolate the unknown number ( 87)- having on hand one of the subset's data - no matter what algorithm you apply - you'll get different numbers.

My conclusion - Given 2 sets of integers A of n(A) members and B of n(B)- you can define the bonds for n( intersection set A x B) : generally between 0 and <n(the smaller set).

The result **might be close** to the ratio **n(A)/N(B)** for an uniform distribution in both sets -but for a non-uniform only the upper and lower bonds are known .

Our case is not of **intersection **but of **matching **- not solvable in general case.

*Edited on ***April 18, 2017, 4:16 am**