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Q 9. In this question, you must clearly set out your working, describing your process with full English sentences. (a) Using

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Solution: Q 9.

\small \left ( a \right )

Using Euclid's Algorithm we have,

33 = 1 x 25 +8...... (1)

\small 25=3\times 8+{\color{Red} 1}......\left ( 2 \right )

\small 8=8\times 1+0

The last non-zero remainder is \small 1 and therefore \small \mathrm{gcd}\left ( 25,33 \right )=1.

From \small \left ( 1 \right ) we have,

\small 8=33-1\times 25......\left ( 3 \right )

From  \small \left ( 2 \right ) we get

\small 1=25-3\times 8

\small \therefore 1=25-3\times \left ( 33-25 \right ), from 3

\small \therefore 1=25-3\times 33+3\times 25

\small \therefore 1=4\times 25-3\times 33

\small \therefore x=4,\: \: \: \: y=-3

\small \left ( b \right )

From \small \left ( a \right ) , we have

\small \mathrm{gcd}\left ( 25,33 \right )=1.

\small \therefore 25 has a multiplicative inverse in \small \mathbb{Z}_{33}

Also, \small 1=4\times 25-3\times 33

The value of \small x ( the coefficient of \small 25 ) is \small 4 , so  \small 25^{\left ( -1 \right )}=4 .

Check:  \small 25\times 4=100=3\times 33+1\equiv 1\left ( \mathrm{mod}\: 33 \right )

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