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GATE CSE 2023 | Question: 22

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$\left(132\right)_{4}$

= $(1*4^2 + 3*4^1 + 2*4^0)_{10}$

= $\left(30\right)_{10}$

= $\left(25 + 5 + 0\right)_{10}$ [powers of 5]

= $(1*5^2 + 1*5^1 + 0*5^0)_{10}$

= $\left(110\right)_{5}$

Answer is 110.

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Firstly convert the base $4$ into decimal number system (Base $10$):

$(132)_4=(z)_{10} \implies 2*4^0+3*4^1+1*4^2=(30)_{10}$

Now convert $(30)_{10}$ into base $5$,dividing by $5$ we get:

$(30)_{10}=(x)_5$

$\therefore (x)_5=(110)_5$

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