I'm a little put out by the concept of 1111 in base 1. In base 10 you can use the digits 0-9, in base 3 0-2 and so on. Logically that should lead to only using the digit zero in base 1. If you're going to allow 1, then why stop there? Why not write 2323 for 3*1^0 plus 2*1^2 plus 3*1^3 plus 2*1^4 = 1111111111 (or 10 decimal)?

In the monary numeration system the zero is just a non existing entity, it was an empty space. There is evidence that primitive cavemen carved in a stickbones grooves to tally their hunting pieces. They leaved an empty space probably for the no hunting period and started to carve again.

The zero was a creation of the Babylonian, they used just as a placeholder, otherwise without it their numeration system based on position could produce different numbers, so YY could mean 60 or 3601 so they decided to write YY for 60 and Y#Y for 3601 with the spaceholder # (obviously my keyboard has not same symbols the babylonian used Y for 60 or # for the zero).

But unfortunally the zero got lost in the Western Civilization, until was introduced by Fibonacci in the thirteen century, and it took almost two centuries to get accepted.

Romans or Greeks numerals had no symbol for zero,they did not need a placeholder. So they start to count the calendar with year one, that is why we got into the discussion if 2000 or 2001 was the first year of the twenty one century.

Probably you can get more accurate information but that is what I remember.

Yes, I've always been a bit bothered by that. It'seems to me that you can't extend the standard interpretation of base n notation to the case n=1. You should only be allowed one digit'symbol, and it should be 0, but in that case, the only number you can represent is 0 itself, since, for instance,000 = 0*1^2 + 0*1^1 + 0*1^0 = 0+0+0 = 0.

On the other hand, using 1 as the digit does provide a unique string of digits to name any positive integer, so this version of base 1 does function as a numeration system. I see two possibilities here: first, there may be some non-standard way of looking at positional notation that agrees with the standard way for n>1 and also works for the n=1 case. If so, I don't know what it might be -- perhaps it is an interesting puzzle.

Second, perhaps one shouldn't look at these strings of "1"s as a base system at all. After all, since 1^n = 1, all the positions have equivalent values, so that it isn't really "positional" after all, since the position of a digit has no bearing on its value. Instead, it is a primitive counting notation where the symbols are in direct one to one correspondence with the items counted, like cutting notches in a stick, or using hashmarks to keep a tally.

I think that's it -- base 1 is really just hashmarks, which happen to look like the numeral "1" (which probably started out as just a hashmark itself, so it's no coincidence), and therefore got to be called "base 1" but are really not related to the base n positional system at all.

--Stuart Anderson

First I try to solve without move anything so I set

(B^0)+(B^1+B^0)+(B^2+B^1+B^0) = (B^3+B^2+B^1+B^0) or simplifying:

B^3+B^2-B^1=2 is a cubic without an inmmediate solution, so I realize you must swicth some 1 and mess with the sign to get a more easy root for B the unknown base.

The easiest but not the only one I found is:

(B^0)+(B^0)+(B^0)+(B^0)= (B^3+B^2+B^1+B^0)

what yield B=1. 0r 1+1+1+1 = 4.

I believe the original post said that you could not change the right side of the equation.

--Stuart Anderson