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Question IN AN EXAM : $(22)_4$ + $(101)_3$ – $(20)_5$ = $(x)_4$ + $(4)_{x–1}$. The value of $x$ is ___________.

According to me the question is invalid, since the exam did not specify the base of the answer...

reason :

Unless you specify which "$x$" ,

$x$ = $12_4$ and $x$ = $6_{10}$ .

For $4_{x-1}$ ,logically, the value of $x-1$ ranges from $5$ to $infinity$ with no change

($4_{100}$ == $4_{11}$ == $4_5$ == $4_{9999}$ == $4_{16}$)

BUT

since we "always" denote a BASE/RADIX always in "to the BASE 10" form

For example we denote a hexadecimal number in the following form

<p>$34CF_{16}$ and not $34CF_{F1}$ [$16$ in Hexadecimal is $F1$] or</p>

<p>$34CF_{10000}$ [16 is 10000 in binary] and vice versa)</p>

<p>the value of $x-1$ is definitely $(5)_{10}$ since in $4_{x-1}, x-1$ is a BASE</p>

so $x-1_{10}$ = $5_{10}$

So the answer is

$x = 12$ ( for the question $(x)_4$ )

$x - 1 = 5$ ,Thus $x=6$ ( for the question $(4)_{x-1}$ )

Is my logic correct?

Paul
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  • I think the question is valid. x must be a string of digits that in base 4 + 4 = the value of the lhs. I.e. x = 12. No ambiguity so for as I can tell. – fleablood Oct 08 '16 at 02:21

1 Answers1

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Convert the LHS to decimal, which gives $10 + 10 -10 = 10$

On the RHS $4$ to any base would be $4$ decimal, since it's just a single digit.

So you're left with $x_4 = 6_{10}$, giving $x = 12$.

The $4$ is just a single digit, and is unchanged in value in any (valid) base, as mentioned, so there is no chance at all for ambiguity. But even otherwise, you should treat the $x$ as a literal digit representation (or a "string of digits" as @fleablood put it). Substituting back gives the initial equation: $22_4$ + $101_3$ – $20_5$ = $12_4$ + $4_{11}$ which is what was intended.

Deepak
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