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aaabaaa and perfect square (Posted on 2010-07-10) Difficulty: 3 of 5
Determine all possible 7-digit perfect square(s) (in base 8) having the form aaabaaa, where a is a digit from 1 to 7 and b is a different digit from 0 to 7.

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

No Solution Yet Submitted by K Sengupta    
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Solution solution (partial analytical deduction) Comment 1 of 1

The last digit of a perfect square in base 8 must be 0, 1 or 4. As it is given that a can not be 0 ("...a is a digit from 1 to 7..."), a must be 1 or 4. Thus, we know aaabaaa must be 111b111 or 444b444, reducing the 49 possible numbers to 14.

Converting 11101118 into base 10 and finding its digital root we have 299081 with a digital root of 2. As the digital root of 512 (8^3) is 8, the digital root will decrement by 1 (9-8=1; skipping 0 as digital roots range from 1 to 9) for each possible value of b, e.g., 11111118 will have a digital root of 1, 11121118 will have a digital root of 9, etc.  As the digital root of a perfect square in base 10 must be 1, 4, 7 or 9, the possible squares would be 11111118 (eliminated as a and b must be different), 11121118 , 11141118 , 11171118. Performing the same task for 4440444= 1196324; base 10 digital root = 8. Possible squares are then 44414448, 44454448, and 44474448.
Converting these six numbers into base 10 and taking the square root of each we find only one perfect square: 44414448.

11121118 = 300105:Digital Root=9:Sq. Root=~ 547.818
11141118 = 301129:Digital Root=7:Sq. Root=~ 548.752
11171118 = 302665:Digital Root=4:Sq. Root=~ 550.150
44414448 =1196836:Digital Root=7:Sq. Root=  1094 <<<
44454448 =1198884:Digital Root=4:Sq. Root=~ 1094.94
44474448 =1199908:Digital Root=1:Sq. Root=~ 1095.40

 

Edited on July 10, 2010, 3:54 pm
  Posted by Dej Mar on 2010-07-10 13:41:36

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