i am no understanding how to compose numbers in bases. Because that example, \$10=10 imes11^0\$ for this reason it must be \$10\$ in base \$11\$. But additionally \$11=0 imes11^0+1 imes11^1\$ therefore \$11\$ is \$10\$ in basic \$11\$. For this reason both \$10\$ and also \$11\$ are \$10\$ in basic \$11\$? 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.