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 1110111_{8} 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 (98=1; skipping 0 as digital roots range from 1 to 9) for each possible value of b, e.g., 1111111_{8} will have a digital root of 1, 1112111_{8} 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 1111111_{8} (eliminated as a and b must be different), 1112111_{8 }, 1114111_{8 }, 1117111_{8}. Performing the same task for 4440444_{8 }= 1196324; base 10 digital root = 8. Possible squares are then 4441444_{8}, 4445444_{8}, and 4447444_{8}.
Converting these six numbers into base 10 and taking the square root of each we find only one perfect square: 4441444_{8}.
1112111_{8} = 300105:Digital Root=9:Sq. Root=~ 547.818
1114111_{8} = 301129:Digital Root=7:Sq. Root=~ 548.752
1117111_{8} = 302665:Digital Root=4:Sq. Root=~ 550.150
4441444_{8} =1196836:Digital Root=7:Sq. Root= 1094 <<<
4445444_{8} =1198884:Digital Root=4:Sq. Root=~ 1094.94
4447444_{8} =1199908:Digital Root=1:Sq. Root=~ 1095.40
Edited on July 10, 2010, 3:54 pm

Posted by Dej Mar
on 20100710 13:41:36 