who can help me question 4 and 5 Find the smallest positive integer k such that

Question

who can help me question 4 and 5 Find the smallest positive integer k such that 16 middot 5^32 + 89 + k is exactly divisible by 6. Find the multiplicative inverses, if they exist for: (i) 12 mod 15 (ii) 12 mod 13 Find the solution, if any exist, of: (i) 7x 3 mod 9 (ii) 17x 4 mod 36 You have till Friday (8/04) 4:00pm to submit this assignment. The assignment should be submitted in the assignment box marked with the course number (out- the CS Learning Centre), anytime after Wednesday. Please. DO NOT LEAVE AT MY OFFICE.

Explanation / Answer

4.

i)

Let x be multiplicative inverse of 12 mod 15

Hence, 12x=1 mod 15

or 12x-1=0 mod 15

ie 12x-1 is a multiple of 15

12x-1=15k

But 3 |12 and 3|15 so we must have: 3|1 which is not true. Hence a contradiction

So multiplicative inverse of 12 mod 15 does not exist.

ii)

12 mod 13

Note: 12=13-1 =-1 mod 13

(-1)^2=1 mod 13

Hence, 12 is its own multiplicative inverse.

Check:12^2=144=143+1=13*11+1=1 mod 13

5.

i)

7x=3 mod 9

7=9-2 =-2 mod 9

So we can rewrite equation as:

-2x=3 mod 9

2x=-3 mod 9

2x=9-3 mod 9

2x=6 mod 9

So we need to check which multiples of 2 are: 6 mod 9

6 mod 9 numbers are:6,15,24

So , 6 is the solution as:24=2*12=2*(9+3)=2*3 mod 9

So, x=3 mod 9 is the solution

2*3=6 mod 9

ii)

17x=4 mod 36

36=9*4

So we must have:

17x=4 mod 9, 17x=4 mod 4

17x=4 mod 9 gives:

(18-1)x=4 mod 9

-x=4 mod 9

x=-4 mod 9

Now lets look at other equation

17x=4 mod 4

(16+1)x=4 mod 4

x=4 mod 4

Or ,x=0 mod 4

So we need to find x so that:

x=0 mod 4, x=-4 mod 9

x=-4 mod 9=5 mod 9

x=5 mod 9 gives:

5,14,23,32

Only x so that:x=0 mod 4 is:x=32

So, x=32 is the solution.

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