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Pandigital and square (Posted on 2017-04-15) Difficulty: 3 of 5
There are 87 square numbers that use every digit (base 10, no leading zeroes)) exactly once.

List them.

No Solution Yet Submitted by Ady TZIDON    
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re(4): Pencil and Paper solution (spoiler) Comment 6 of 6 |
(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
  Posted by Ady TZIDON on 2017-04-18 02:08:12

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