In base 10, we have 10 signs for digits. In basic 2, we have 2. Likewise, in base 11, us must have 11 - however 0, 1, 2, ..., 9 just makes ten. So we need to introduce a new symbol (we typically use A) to be the eleventh digit. "A" represents a worth of 10 in a solitary digit.

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So in base 11, 10 will certainly be denoted A (because it is A 1"s and also zero 11"s). 11 will certainly be denoted 10 (because it is one 11 and also zero 1"s). Similarly, 21 would certainly be 1A (one 11 and A 1"s).

For one integer base $b ge 2$ and nonnegative integer $n$, we can express $n$ in basic $b$ using the notation $$n = (d_m, d_m-1, ldots, d_1, d_0)_b = sum_k=0^m d_k b^k,$$ where $$d_k in , 1, ldots, b-1$$ for all $k$; moreover, together a depiction is

**unique**. The is to say, writing $n$ in basic $b$ amounts to finding a succession $d_k\_k ge 0$ of nonnegative integers

**strictly less than**$b$ because that which the over sum equals $n$.

For example, allow us select $n = 32597$, $b = 23$. Climate the largest power the $23$ the is much less than or same to $n$ is $23^3 = 12167$, and also the the quotient that $n$ split by $23^3$ is $2$. This tells us $d_3 = 2$, for this reason $$n = 2 cdot 23^3 + d_2 cdot 23^2 + d_1 cdot 23^1 + d_0 cdot 23^0.$$ To solve for the subsequent terms, we simply subtract and proceed v the remainder: $$n - 2 cdot 23^3 = 8263,$$ and the largest power that $23$ that is much less than $8263$ is $23^2 = 529$, and also $lfloor 8263 / 529 floor = 15$, for this reason $d_2 = 15$. Repeating the process until the end, we achieve $$32597 = 2 cdot 23^3 + 15 cdot 23^2 + 14 cdot 23^1 + 6 cdot 23^0,$$ and we deserve to write this more compactly together $$n = (2, 15, 14, 6)_23.$$

This to be a fairly detailed example. In ~ this point, it"s worth noting that, for convenience, we have the right to think the such representations as comprising an **infinite** succession of digits $d_k$, and that if $n$ is finite, then there exists some $m$ for which $d_k = 0$ for all $k > m$; so for example, $$(2,15,14,6)_23 = (ldots, 0, 2, 15, 14, 6)_23$$ or if we wrote the digits in "reverse"--that is, in order of boosting significance--we could say $$d_k\_kge 0 = (6, 14, 15, 2, 0, 0, 0, ldots).$$ This affords united state the convenience that not having to point out $m$, and also that in part sense, the base-$b$ representation of any type of nonnegative essence is an infinite sequence that nonnegative integers much less than $b$.

That said, exactly how would us answer the original question? Well, in base $b = 11$, the number $n = 10$ is much less than $b$, for this reason we just write $$10 = (10)_23 = 10 cdot 11^0.$$ Thus, the sequence equivalent to together $n$ and $b$ is $$d_k\_kge 0 = (10,0,0,0,ldots).$$ note we do not need to resort to using letters or other symbols.

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What is $11$ in basic $11$? Well, this have to be clear now: $$11 = 1 cdot 11^1 + 0 cdot 11^0 = (1,0)_11 = (0,1,0,0,0,ldots).$$

Let"s try some more. What is $n = 57$ in basic $b = 3$? We have actually $$57 = 2 cdot 3^3 + 0 cdot 3^2 + 1 cdot 3^1 + 0 cdot 3^0 = (2,0,1,0)_3 = (0,1,0,2,0,0,0,ldots).$$ How about $300$ in base $100$? This is just $$300 = 3 cdot 100^1 + 0 cdot 100^0 = (3,0)_100 = (0,3,0,0,0,ldots).$$ by now, girlfriend should discover the process familiar. Try more examples; simply pick an integer $b ge 2$ and also any nonnegative $n$, and work it out yourself.