Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

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?


Grade:

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

Please feel free to post as many doubts on our discussion forum as you can. If you find any question 
Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We 
are all IITians and here to help you in your IIT JEE preparation. 


All the best s adhithya !!!




Regards,

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

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free