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If two natural numbers, x and y are selected at random, then the total number of ways such that x 2 + y 2 is divisible by 10 is?

If two natural numbers, x and y are selected at random, then the total number of ways such
that x2 + y2 is divisible by 10 is?


2 Answers

Askiitians Expert Sahil Arora - IITD
19 Points
11 years ago

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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Askiitians Experts

Sahil Arora

mycroft holmes
272 Points
11 years ago

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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