How to convert $(FADE)_{16}$ to base 11 easily? Is there any shortcut method?
My attempt -
First I converted base 16 to base 10 then divided with 11 to get answer. But this method was complex and time consuming.
First on converting in decimal I got 64222.
Then on dividing by 11 consecutively I got 44284.
Converting to/from base $10$ is easy because you know the powers of $10$ well and can do arithmetic in base 10 quickly. To convert directly from one base to another, it helps to know the powers of the target base expressed in the starting base and it helps to be able to do arithmetic in the starting base quickly. In your case, you want to know the powers of $11$, expressed in base $16$. \begin{align*} 11_{10}^0 &= 10_{11}^0 = 1_{16} \\ 11_{10}^1 &= 10_{11}^1 = \mathrm{B}_{16} \\ 11_{10}^2 &= 10_{11}^2 = 79_{16} \\ 11_{10}^3 &= 10_{11}^3 = 533_{16} \\ 11_{10}^4 &= 10_{11}^4 = 3931_{16} \\ 11_{10}^5 &= 10_{11}^5 = 2751\mathrm{B}_{16} \end{align*} This last is larger than $\mathrm{FADE}_{16}$, so we know that we only need five digits base $11$.
Dividing $3931_{16}$ into $\mathrm{FADE}_{16}$, we get $4_{16}$ with remainder $161\mathrm{A}_{16}$.
Dividing $533_{16}$ into $161\mathrm{A}_{16}$, we get $4_{16}$ with remainder $14\mathrm{E}_{16}$.
Dividing $79_{16}$ into $14\mathrm{E}_{16}$, we get $2_{16}$ with remainder $5\mathrm{C}_{16}$.
Dividing $\mathrm{B}_{16}$ into $5\mathrm{C}_{16}$, we get $8_{16}$ with remainder $4_{16}$.
So now we know \begin{align*} \mathrm{FADE}_{16} &= 4_{16} \cdot 10_{11}^4 + 4_{16} \cdot 10_{11}^3 + 2_{16} \cdot 10_{11}^2 + 8_{16} \cdot 10_{11}^1 + 4_{16} \cdot 10_{11}^0 \\ &= 4_{11} \cdot 10_{11}^4 + 4_{11} \cdot 10_{11}^3 + 2_{11} \cdot 10_{11}^2 + 8_{11} \cdot 10_{11}^1 + 4_{11} \cdot 10_{11}^0 \\ &= 44284_{11} \end{align*}
Note that in the "convert the individual digits from base $16$ to base $11$" step there can be carries from one position to the position to the left, although this particular problem doesn't need it. For instance, if we had $\mathrm{C}_{16}$ for the middle digit (instead of $2_{16}$), we would have $\mathrm{C}_{16} = 11_{11}$, so the result would have been $45184_{11}$.
If you haven't practiced with hexadecimal, you are probably performing the divisions above by converting to decimal and back, so you aren't gaining anything by this method. If you have practiced, then you can do the above and convert to base $11$ pretty much directly.
$\begin{array}{crcrcc} 1.&FADE_{16}/B_{16}&=&16CE_{16}&\text{remainder}&4_{16}\\ 2.&16CE_{16}/B_{16}&=&0212_{16}&\text{remainder}&8_{16}\\ 3.&0212_{16}/B_{16}&=&0030_{16}&\text{remainder}&2_{16}\\ 4.&0030_{16}/B_{16}&=&0004_{16}&\text{remainder}&4_{16}\\ 5.&0004_{16}/B_{16}&=&0000_{16}&\text{remainder}&4_{16} \end{array}$
$44284_{11}$
If a result digit ever turns out to be 10, be prepared with a character to stand in for it, since you are wanting an answer in undecimal (base 11).
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Hey, Buddy, take my advice and do a one-time conversion to decimal (base 10). Believe me, doing arithmetic in an unfamiliar base is a pain in the derrière.
$FADE_{16} = 64222_{10}$
$\begin{array}{crcrcc} 1.&64222_{10}/11_{10}&=&5838_{10}&\text{remainder}&4_{10}\\ 2.&5838_{10}/11_{10}&=&530_{10}&\text{remainder}&8_{10}\\ 3.&530_{10}/11_{10}&=&48_{10}&\text{remainder}&2_{10}\\ 4.&48_{10}/11_{10}&=&4_{10}&\text{remainder}&4_{10}\\ 5.&4_{10}/11_{10}&=&0_{10}&\text{remainder}&4_{10} \end{array}$
$44284_{11}$
Your example, while valid, is not a very good teaching example, since all of your result digits have the same representations in decimal, undecimal and hexadecimal (bases 10, 11 and 16).
If a result digit ever turns out to be 10, be prepared with a character to stand in for it, since you are wanting an answer in undecimal (base 11).
