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 Subtract 1 from squared abbbb (Posted on 2008-08-16)
Determine all possible five digit positive decimal (base 10) integer(s) of the form abbbb, with a ≠ b, that contain no leading zeroes, such that (abbbb2 - 1) is equal to a positive ten digit integer (with no leading zeroes) containing each of the digits 0 to 9 exactly once.

Note: While the solution may be trivial with the aid of a computer program, show how to derive it without one.

 See The Solution Submitted by K Sengupta Rating: 4.0000 (1 votes)

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 analytical and brief | Comment 3 of 6 |

*   N is over 3222

**  bb cannot be 33 66 or 99 since we get repetition of digits  within the last three places

*** N^2= 1mod 9 so N= 1mod 9  or  -1mod 9

****  a<>b        ........not equal

So only ten candidates remain:

61111 41111  92222 34444 14444
85555 65555 97777   38888  and 58888

Out of those only 8555 and  9777  qualify.

 Posted by Ady TZIDON on 2008-08-17 10:24:23

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