Discrete structures Example for number 3 and 4 and if you ca

Solution

1 ) a) 33 mod 6

33 = 6 *5+ 3 ,so remainder is 3

hence 33 mod 6 = 3

b) 54 mod 13

54 = 13*4 +2

hence  54 mod 13 = 2

c) 0 mod n

clearly remainde will be 0.

2) n = dq + r

n = -101 , d = 5

-101 = 5 q+ r

-101 can be written as 5 * (-21) +4

hence r = 4 , q = -21

3) x^2 + x +5

= x* (x +1) +5

now for number x , x*(x+1) will be even ,because it the product of two consecutive number,

so adding 5 to even number will be odd.

4) x*(x2 -1) *(x+2)

= (x-1)*x*(x+1)*(x+2)

it is the product of 4 consecutive number , so

there will 2 even numbers and 2 odd numbers

so their product will be multiple of 4.

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