Yep, and to brhan: there are number bases based off of irrational numbers, but they're kind of complicated lol. My favorite is phinary, just cuz the Golden Ratio (and fibonacci numbers) are so amazing

anyway, how I would

*mathematically* solve #3 if I didnt already know the answer:

10101 in base x is, in other words:

x

^{0} + x

^{2} + x

^{4} in base 10

or

1 + x

^{2} + x

^{4} in base 10

changing 273 from base x+6 into base 10 is also easy:

2*(x+6)

^{log(100)} + 7*(x+6)

^{log(10)} + 3*(x+6)

^{log(1)}or:

2*(x+6)

^{2} + 7*(x+6)

^{1} + 3

now that they are both in base 10, they can equal each other:

1 + x

^{2} + x

^{4} = 2*(x+6)

^{2} + 7*(x+6)

^{1} + 3

subtract 1 from both sides

x

^{2} + x

^{4} = 2*(x+6)

^{2} + 7*(x+6)

^{1} + 2

(x+6)

^{2} is (x+6)(x+6), which is x

^{2}+12x+36, all multiplied by 2 is 2x

^{2}+24x+72

7*(x+6)

^{1} is 7x+42

x

^{2} + x

^{4} = 2x

^{2}+24x+72 + 7x + 42 + 2

now we have a ton of like terms on the right side to merge:

x

^{2} + x

^{4} = 2x

^{2} + 31x + 116

subtract x

^{2} from both sides

x

^{4} = x

^{2} + 31x + 116

subtract x

^{4}0 = -x

^{4} + x

^{2} + 31x + 116

*-1

0 = x

^{4} - x

^{2} - 31x - 116

factor

0 = (x-4)(x

^{3}+4x

^{2}+12x+17)

From there, what can easily turn that into 0. The clear answer is 4, cuz 4-4=0, which would negate the rest of it. If you try that, 10101 in base 4 is 273 in base 10 (4+6)

I believe this was brhan's method. This is the mathematical/algebraic way to do it.

The logic way to do it, which is probably what I would have really done, is:

* x+6 cant be 7 or smaller, cuz a 7 appears in the number so it would carry over as a 10 if x+6 was 7 or lower. Therefore x+6 is 8 or more

* which means x is 2 or more

* x+6 could be more than 10, but no symbols were used in 273 so we're not sure if x+6>10 yet, though it's unlikely, since no symbols were used in 273, so it's not showing the part of the base, and what symbols would you use, letters..? To see if that's an option, figure out if x was higher than 4 and see if that's a possible solution. However if x was higher than 4, for example 5, than 1+5^2+5^4 = 651 in base 10, or, in base 11, 542, already much too big than 273

* so our range is x=2 or 3 or 4

* if x was 2, simple binary, then x^4+x^2+x^0 is 16+4+1 = 21 in base 10, which is (16=20)+5 = 25 in base 8 (8=x+6 if was x was 2). Clearly 25 is much less than 273.

* I could try x=3, but I can already see that 1+9 is 10 in base 10, or 11 in base 9, so 3^4 would have to be 262 in base 9, but I know that 3^4 is 9^2 and 9^2 is 81 in base 10- which is very cleanly 100 in base 9. 100+11=111 in base 9, not 273. Not even close. So I dont even need to check base 3 fully

* the only thing left is x=4

* 1+16+256 is 273. Yay! The number! But wait- dont you have to convert it to base 10, so it will change *sinking feeling* But not! Because if x is 4, than x+6 is 10. 273 is in both base 10 and base x+6.

Therefore x=4