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Add 1, get a square (Posted on 2008-07-17) Difficulty: 3 of 5
By adding 1 to the non leading zeros number N = AAABBBCCC, it becomes a square. Find N.

A, B, and C are different digits.

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

See The Solution Submitted by pcbouhid    
Rating: 5.0000 (1 votes)

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Solution Solution (partial analysis) | Comment 2 of 4 |
Adding 1 to the number N will give us a pattern of AAABBBCCD (C < 9) or AAABBD000 (C = 9). As all squared integers end in one of the six digits: 0, 1, 4, 5, 6 or 9; C must be either 9, 0, 3, 4, 5, or 8. This limits us to the following patterns for our possibile square:
  • AAABBB001
  • AAABBB334
  • AAABBB445
  • AAABBB556
  • AAABBB889
  • AAABBD000
For any perfect square that is an even number, it must be a composite with an even number of the prime factor 2, i.e, it must be divisible by 4. As AAABBB000 must be divisible by 4, we only need to check the last three digits of AAABBB334. 334 modulo 4 is 2 and not 0, thus, AAABBB334 cannot be a perfect square, and may be excluded. Hence, C cannot equal 3.

For any perfect square with its ones-digit as a 5, its tens-digit must be a 2. This isn't the case for AAABBB445, therefore we can exclude AAABBB445. Hence, C cannot equal 4.

As any perfect square with any trailing zeroes must be a multiple of 10, it must also have an even number of trailing zeroes. If D were to be 0, B would also have to be 9; which cannot be, as given, B must be distinct from C. Thus, we can exclude AAABBD000. Hence, C cannot equal 9.

As A cannot be 0, and with A, B and C being distinct, this gives us a total of 200 possible numbers to check. Only one of these numbers is a perfect square: 111999889 = 105832. Thus, our number must be 111999889 - 1 = 111999888.
  Posted by Dej Mar on 2008-07-17 20:26:49
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