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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.

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    $\begingroup$ It would be most helpful when you include the answer you got, given your efforts. I'm asking because your description "... then divided by 11 to get the answer" is entirely unclear. So I simply ask you to edit your post to include the result you obtained. That will help us better in targeting an answer specifically for you. $\endgroup$ – Namaste Sep 28 '17 at 17:14
  • $\begingroup$ @amWhy please see now. $\endgroup$ – Amar Sep 28 '17 at 17:20
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    $\begingroup$ Sure you could bypass converting to base 10 if you were comfortable working in base 16 or base 11. Doing, say, long division in base 10 is just a set of rules that seem easy due to years of practice. But there’s nothing (in principle) stopping you from following the equivalent rules in base 16 :). Or conversely, performing the multiplication in base 11. $\endgroup$ – Erick Wong Sep 28 '17 at 17:21
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    $\begingroup$ For example, working in base 11: $F_{16} = 14_{11}$, multiply by $15_{11}$ to get $F0_{16} = 1A9_{11}$, add $A_{11}$ to get $FA_{16} = 208_{11}$, etc., etc. $\endgroup$ – Daniel Schepler Sep 28 '17 at 17:27
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    $\begingroup$ @Amar, of course it was. Some folks, growing up with base 10 initiation, move on to master base 2, base 8, base 16, etc. I'm just saying that base 11 is not as familiar as base 2^n (e.g., base 2^4 = 16), or base 10. That's all. It can certainly be done. What I'm asking is about what you mean by "shortcut" method? I.e., If you are looking for a "speedier way" the answer is likely "no". But if you are looking for something else, then I don't yet understand what you mean by "shortcut". So you might want to add to your post what you mean by "shortcut". $\endgroup$ – Namaste Sep 28 '17 at 17:40
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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.

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$\begin{array}{crcrcc} 1.&FADE_{16}/B_{16}&=&16CE_{16}&\text{remainder}&4_{16}\\ 2.&16CE_{16}/B_{16}&=&0212_{16}&\text{remainder}&8_{16}\\ 3.&0212_{16}/B_{16}&=&0030_{16}&\text{remainder}&2_{16}\\ 4.&0030_{16}/B_{16}&=&0004_{16}&\text{remainder}&4_{16}\\ 5.&0004_{16}/B_{16}&=&0000_{16}&\text{remainder}&4_{16} \end{array}$

$44284_{11}$

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).

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