Considering k from 0 to 9 and n=10
Case(1): h(k)=k%10
0%10=0,1%10=1,2%10=2 similarly 9%10=9
Collision rate=0
Case(2):h(x)=k*k%10
(0*0)%10=0
(1*1)%10=1
(2*2)%10=4
(3*3)%10=9
(4*4)%10=6
(5*5)%10=5
(6*6)%10=6
(7*7)%10=9
(8*8)%10=4
(9*9)%10=1
Collision rate=4/10
Case(3):h(k)=(gcd(k+1),(2k+2)+k)%10
0--->(gcd(1,2)+0)%10=1
1-->(gcd(2,4)+1)%10=3
2-->(gcd(3,6)+2)%10=5
3-->(gcd(4,8)+3)%10=7
4-->(gcd(5,10)+4)%10=9
5-->(gcd(6,12)+5)%10=1
6-->(gcd(7,14)+6)%10=3
7-->(gcd(8,16)+7)%10=5
8-->(gcd(9,18)+8)%10=7
9-->(gcd(10,20)+9)%10=9
Collision rate=5/10
Since collision rate is highest for case(3), it is least likely to be selected as a hash function