Problem:
If the base 8 representation of a perfect square is ab3c, where aî€ =0, then c is
Answer Choices:
A. 0
B. 1
C. 3
D. 4
E. not uniquely determined
Solution:
If n2=(ab3c)8​, let n=(de)8​. Then n2=(8d+e)2=64d2+8(2de)+e2. Thus, the 3 in ab3c is the first digit (in base 8) of the sum of the eights digit of e2 (in base 8) and the units digit of (2de) (in base 8). The latter is even, so the former is odd. The entire table of base 8 representations of squares of base 8 digits appears below.
ee2​11​24​311​420​531​644​761​​
The eights digit of e2 is odd only if e is 3 or 5; in either case c, which is the units digit of e2, is 1. (In fact, there are three choices for n:(33)8​,(73)8​ and (45)8​. The squares are (1331)8​,(6631)8​ and (2531)8​, respectively.)
Alternate solution sketch (using familiarity with number theory). We are given n2=(ab3c)8​=83a+82b+8⋅3+c.
Since n2≡0,1 or 4mod8, we must have c=0,1 or 4. If c=0, then n2≡ 8(8K+3)mod8, an impossibility since 8 is not a square. If c=4, then n2 =4(8L+7)mod8, another impossibility since no odd squares have the form 8L+7. Thus c=1.