[Guest]

For a positive octal (base 8) perfect square N with no leading zero, determine the probability that N has the form aaabaaa.

Edited February 12, 2010 by K Sengupta

[superprismatic]

The probability is 1/937. There are 937 squares which are comprised of 7 octal digits without a leading 0. Only 1 has the appropriate form and it is 4441444 in octal.

[Guest]

[spoiler='Like superprismatic, we assume that the sample space is the set of 7 octal-digit perfect squares with no

leading zero.']A number N = n² (with n integer) has 7 octal-digits with no leading zero if and only if N is in

[(1000000)₈, (10000000)₈) = [8⁶, 8⁷) = [2¹⁸, 2²¹).

This is equivalent to

n is in [2⁹, 2¹⁰√2) = [512, 1024√2)

or (since n is integer)

n is in [512, 1448].

We conclude that there are exactly 1448 - 512 + 1 = 937 such numbers N.

Now we shall see how many of those numbers have the stated form, that is,

n² = N = a(1 + 8 + 8² + 8⁴ + 8⁵ +8⁶) + b8³ = (1 + 8 + 8²)(1 + 8⁴)a + 512b = 73*4097a + 512b (*)

where a and b belong to {0, ..., 7} and a is non zero.

From (*) we get a = n² (mod 8) and since the only perfect squares (mod 8) are 0, 1 and 4 it follows that a = 1 or a = 4.

Substituting each of the 16 pairs (a, b) in {1, 4}x{0, ..., 7} into (*), the only one which gives a perfect square is (4, 1), in which case, N = (4441444)₈ = 73*4097*4 + 512*1 = 1094².

Therefore, the probability is 1 / 937.