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If two natural numbers, x and y are selected at random, then the total number of ways suchthat x2 + y2 is divisible by 10 is?
Dear s adhithya
IF both x and y are divisible by 10.then obviously
x2 + y2 will also be divisible by 10 ...
now how many natural no. u can find divisible by 10..
obviously infinite......
so there are infinite ways in which we can select x and y
I think the question doesn't make sense..there should be some subset of natural no. given
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Consider the set of possible remainders on division by 10 i.e.{1,2,...,10}. When you square the possible remainders are {0,1,4,5,6,9}.
If we consider the ordered pairs of remainders (x,y), we get 100 possibilities
Thus x2+y2 will be divisible in the following cases:
(1) (0,0) - 1 case
(2) (1,9) - 4 cases
(3) (4,6) - 4 cases
(4) (5,5) - 1 case
(5) (6,4) - 4 cases
(6) (9,1) - 4 cases
making 18 favourable cases out of 100 possible cases giving a probability of 18/100 = 9/50
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