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.

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 4440444₈ = 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