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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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 A start (spoiler) | Comment 1 of 6
a) The square root of 10 is > 3.16, so in order to be 10 digits, ab must be >= 32

b) b cannot be 0, as that will lead to a number ending in 8 zeroes

b) A 10 digit number containing all 10 digits = 0 mod 9, by casting out 9's.

Therefore abbbb2 = 1 (Mod 9)
Therefore abbbb = +1 or -1, mod 9
But abbbb mod 9 = a + 4b mod 9 (casting out 9's again)
so a + 4b = +1 or -1 (mod 9)

c) Therefore, the only ab which might work are:

61, 41, 92, 72, 53, 34, 85, 65, 46, 97, 38, 58, 89
 Posted by Steve Herman on 2008-08-16 15:20:25

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