Please give me help on this question Full answers only If an

Please give me help on this question. Full answers only. If answer is handritten, please post a clear copy of it so it is readable.

Solution

a)

9*5=45=44+1=1 mod 11

So we multiply given equation by 5

You can multiply this equation by any number as long as it is coprime to 11

So, 5*9x=5 mod 11

45x=5 mod 11

x=5 mod 11.

So basically it becomes easy if you can find the multiplicative inverse. Here we had to find for 9 mod 11

b)

10x+23=3 mod 31

10x=-23+3=-20 mod 31

10x=11 mod 31

Again it is easy to see

10*3=30=31-1 (Note that for large number this won\'t be easy and we may have to use Euclid Algorithm or something else. We won\'t be able to see the inverse directly)

So we multiply by 3

30x=33=31+2=2 mod 31

30x=-x=2 mod 31

So,

x=-2 mod 31

x=31-2 =29 mod 31

This is the solution

c)3x+15=4 mod 21

3x=4-15 mod 21

3x=-11 mod 21

Now here x is multplied by 3 so we cannot multply by any number so that ax=1 mod 21 because 3 multiplied by any number is a multple of 3 and 21 is a multiple of 3

Now note 21 is a multiple of 3

3x is a multiple of 3

So 3x and any multiple of 21 always differ by a multiple of 3

But 11 is not a multplie of 3

So no solutions

d)

3x+2y=0 mod 7

2x+y=4 mod 7 ie y=-2x +4mod 7

3x+2y=0 mod 7

We already have an equation ie y in terms of 2x mod 7

So we multply by 4 so make coefficient of y equal to 1 and get

12x+8y=0 mod 7

5x+y=0 mod 7

y=-5x mod 7

y=2x mod 7

ANd we have y=-2x+4 mod 7

So,

2x=-2x+4 mod 7

4x=4 mod 7

SO, x=1 mod 7

So, y=2 mod 7