Two no. a $ b are selected at random from natural nos. find prob. that a2+b2 is divisible by 51) 9/252) 7/183) 11/364) 17/81

Hi, For any Natural Number, you can have either of the 10 digits at units place. From these we can select any 2 digits by 10C1 x 10C1 ways (as repetitions are allowed). So total ways of selecting 2 numbers = 100. Now, lets make the cases of getting 5 when 2 squares are added. (0,5) (0,0) (1,4) (1,9) (4,1) (4,6) (5,0) (5,5) (6,4) (6,9) (9,1) (9,6) Note 2 things: i) for total to be 5 any of a or b can be 4 or 6, so (4,6) and (6,4) both are . included, similarly for other cases. ii) cases of 2,3,7 & 8 are not included because any natural number when Squared . can't give these digits at unit places. Then we have to check, when will we get these digits at unit places. 0 -> 0 1 -> 1,9 4 -> 2,8 5 -> 5 6 -> 4,6 9 -> 3,7. Now, we just have to arrange the numbers. For each of (0,0) (0,5) (5,0) and (5,5), we have just 1possibility, that the units digit of a and b both have either 0 or 5. For all other cases, we have 4 possibilities. Since each other digit individually can be obtained from squaring 2 different digits (like 4 is obtained by squaring either 2 or 8). Adding all such cases, the total favorable cases come out to be -> 1x4 + 4x8= 36 (since total 12 ways are listed, out of which 4 have 1 possibility and remaining 8 have 4 possibilities.) Hence taking probability = 36/100 = 9/25, which is the answer. Thanks