Find the smallest three distinct whole numbers A, B and C such that you can rearrange the digits of A and B to get C^2, the digits of A and C to get B^2, and the digits of B and C to get A^2.
**** Leading zeroes are not allowed.
The solution is:
A B C
42005 50042 74162
A^2 B^2 C^2
1764420025 2504201764 5500002244
Method:
This was a difficult programming challenge as the data structures
for the search got very large, with commensurate execution times.
To start off, it is easy to prove that A, B, and C must be
the same length, and that their squares must
be of twice that length. This then requires that
the range of A, B and C is, for example, 31627 to 99999, not
10000 to 99999, where 31627 ~ sqrt(10^9)
To speed things up, I used the idea of matching SODs to screen
for candidate trios. Also, to buy speed, I arranged arrays in
memory for all SOD(A) and SOD(A^2) and all Inventory(A), (A^2),
where the inventory "inv" is just the number of digits, 0
to 9. I also sorted-out candidate arrays for each possible
SOD (e.g., making ~ 90 = max(SOD(A^2)) arrays in the 5 digit case,
since a 5 digit number squared may have an SOD ranging
approximately from 1 to 90. I also included a reverse index
to the A available for each SOD(A^2).
The steps were:
step 1 - check SOD(A&B) = SOD(C^2)
then... if inv(A&B) = inv(C^2)
step 2 - check SOD(A&C) = SOD(B^2)
then... if inv(A&C) = inv(B^2)
step 3 - check SOD(B&C) = SOD(A^2)
then... if inv(B&C) = inv(A^2)
Success!
I ran the search (with custom codes) for the 2, 3, 4 and 5 digit
possibilities. Code is here for
2,
3,
4,
5 digits.
I didn't get a bite until 4 digits, but the sole
4-digit successful trio was degenerate (non-distinct):
3386 9146 9146: 11464996 83649316, and thus incorrect.
Programming 5 digits got hard: I could not make INV(S&B)
with all A concatenated with all B because 10^5 x 10^5 = 10GBy
which leads to segmentation faults in addressing. So I traded
speed for size and had to concatenate inventories on-the-fly.
The 5 digit case needed 30 hrs of a modest CPU's time. Fortunately
the solution popped out in a few hours with A being small compared
9999. Also for the 5 digit case A^2 would only store in the
integer(kind=int64) of the ISO Fortran Environment. With these
hurdles jumped, the solution was reached.
At 5 digits I got the real hits - the solution above,
and another, very nearby, but slightly larger successful trio:
42005 50420 74162 (1764420025 2542176400 5500002244)
where B changed from 50042 to 50420
I also got a degenerate one like the 4 digit one: 33860 91460
and a new degenerate one: 42005 42005 74162.
and another degenerate trio:
42005 74162 42005... Note - these solutions
have the same inv's; they are just
nearly anagrams in general.
inv(A^2)=
0:(2 or 4) 1:0 2:2 3:0 4:2 5:2 6:0 7:0 8:0 9:0
There are only four solutions for 5 digits, listed below.
(There are no solution for 1,2,3, or 4 digits)
42005 50042 74162
42005 50420 74162
48681 63126 93879
50042 50420 74162
For 6 digits, each 5 digit solution x 10 also works. E.g:
420050 500420 741620 is also a solution.
Edited on December 20, 2023, 4:55 am
soln
