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InterviewSolution
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Two natural numbers x and y are chosen at random. What is the probability that `x^(2) + y^(2)` is divisible by 5? |
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Answer» Let r and s be the remainders when x and y are divided by 5, so `x = 5p + r, y = 5q + s,p,q, in N, 0 le r le4, 0lesle4` `therefore x^(2) + y^(2) = 25(p^(2) + q^(2))+ 10(pr+qs)+r^(2)+s^(2) = 5k + (r^(2) + s^(2)), k in N` Thus, `x^(2) + y^(2)` will be divisible by 5 if `r^(2) + s^(2)` is divisible by 5. Therefore, total number of ways is equal to the number of ways of selecting r and s. which is `5^(2) = 25` For favorable ways, we should have `r^(2) + s^(2)` is divisible by 5 and `0 le r^(2) + s^(2) le 32 so r^(2) + s^(2) = 0, 5, 10, 20, 25, or 30`. Thus `{:(r,0,1,2,3,4),(s,0,"2,3","1,4","1,4","2,3"):}` So number of favourable ways is 9. Hence, the probability is 9/25. |
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