In this problem.
a % b means the mod function (i.e residue when a is divided by b).
a/b means integer division (i.e quotient when a is divided by b)
i) (101,22):
101% 10 ≤ 22% 10
1≤ 2 which is true.
(101/10, 22/10)
(10, 2)∊ P need to check is (10, 2) ∊ P
10% 10 ≤ 2% 10
0 ≤2 which is True.
Then (10/10, 2/10)= (1, 0) fails since . (101, 22) not belongs to P
(ii) (22, 101)
22% 10 ≤ 101% 10
2 ≤ 1 is False.
(22, 101) not belongs to P
(iii) (145, 265)
145% 10 ≤ 265 % 10
5 ≤ 5 is true and (145/10, 265/10) = (14, 26) ∊ P has to be checked. Now consider (14, 26).
14% 10 ≤ 26% 10
4 ≤ 6 is true and (14/ 10, 26/10)= (1, 2) ∊ P has to be checked. Now consider (1, 2) 1% 10 ≤ 2% 10 15 2 is true and (1/10, 2/10)=(0, 0) ∊ P which is given to be true. Therefore (145, 265) ∊ P
(iv) (0, 153):
0% 10 ≤ 153% 10⇒ 0 ≤ 3 is true
Then (0/10, 153/10)= (0, 15) should be in P (0,15):
0% 10 ≤ 15% 100 ≤ 5 is true.
Then (0/10, 15/10)= (0, 1) should be in P
(0, 1): 0% 10 ≤ 1% 100 ≤ 1 is true.
Then (0/10, 1/10)= (0, 0) should be in P It is given that (0, 0) ∊ P
Therefore (0, 153) ∊ P So (ii) and (iv) are contained in P