GO Classes DPP | GATE CS | Digital Logic | Base Conversion

Answer 1

First convert each number into decimal form.

$$(132)_r=r^2+3r+2$$

$$(45)_r=4r+5$$

$$(221)_r=2r^2+2r+1$$

Now use the given equation:

$$(132)_r+(45)_r=(221)_r$$

$$\Rightarrow r^2+3r+2+4r+5=2r^2+2r+1$$

Simplify the left side:

$$\Rightarrow r^2+7r+7=2r^2+2r+1$$

Bring all terms to one side:

$$\Rightarrow 0=2r^2+2r+1-r^2-7r-7$$

$$\Rightarrow 0=r^2-5r-6$$

$$\Rightarrow r^2-5r-6=0$$

Factorize:

$$\Rightarrow (r-6)(r+1)=0$$

$$\Rightarrow r=6 \quad \text{ or } \quad r=-1$$

Since a base cannot be negative, $r=6$.

Now check digit validity.

The largest digit used is $5$, so the base must be greater than $5$.

Since $r=6$, it is valid.

Final Answer: $\boxed{r=6}$

Answer 2

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So the answer  r= 6

Answer 3

(132)r​ + (45)r​ = (221)r

$$ 2 * r^0 + 3 * r^1 + 1 * r^2 + 5 * r^0 + 4 * r^1 = 1 * r^0 + 2 * r^1 + 2 * r^2 $$

$$ r^2 + 7r + 7 = 2r^2 + 2r + 1 $$

$$ r^2 - 5r -6 = 0 $$

$$ r^2-6r+r-6 = 0 $$

$$ r(r-6) +1(r-6) = 0 $$

$$ (r+1)(r-6)=0 $$

$$ r=6 (r > 0   \text{ \text{and}   r > \text{All the digits}) $$ 

Answer: r = 6