10 in base 10 is 1010 in base 2. Can you think of any other pairs (n, b) where n in base 10 is nn in base b?
Show this thread
Replacing n with (10x + y), we have: 10x + y = xb^3 + yb^2 + xb + y From which it is clear that for y = 0, (10, 2) is the only solution, since the above reduces to: 10x = xb^3 + xb 10 = b^3 + b Which is only true for b = 2.
-
-
Similarly, for x = 1, we get 10 = b^3 + yb^2 + b Which only holds for b = 2 and y = 0, i.e. (10, 2) again.
Show this thread -
(Actually, b = 1, y = 8 is also a solution -- but an implicit requirement here is that 0 <= x <= y < b, e.g. 8 is not a valid digit in base 1.)
Show this thread -
Anyway, we can rewrite the first expression like so: 10x + y = xb^3 + yb^2 + xb + y 10 = b^3 + (y/x)b^2 + b (y/x)b^2 = 10 - (b^3 + b) But 2 is the only value of b for which b^3 + b <= 10. Thus I have to conclude that (10, 2) is in fact the only solution.
Show this thread -
However, if we generalize base 10 to base c, there are an infinite number of solutions. The expression becomes: c = b^3 + (y/x)b^3 + b With y = 0, there are obviously infinite solutions to c = b^3 + b, e.g. 20_30 = 2020_3. Or with x = 1, y = 1, 11_84 = 1111_4.
Show this thread
End of conversation
New conversation -
Loading seems to be taking a while.
Twitter may be over capacity or experiencing a momentary hiccup. Try again or visit Twitter Status for more information.