Question

A set of 30 consecutive natural numbers is divided into two subsets, A and B. In Set A there are 15 consecutive natural numbers whereas the rest of the consecutive natural numbers are in set B. If the average of the numbers in Set B is 83, What is the last number in Set A? *

Answer 1

Answer:

(a  

2

−b  

2

) is divisible by 3

i.e  (a−b)(a+b) is divisible by 3.

Case(1)  (a+b)  is divisible by 3

If a is chosen from (3k+1) set ,then b must be from (3k+2) set and viceversa.

If a is chosen from (3k) set, then b must be from (3k).

So, number of ways =  10*10

Note that when a is from 3k and b is from 3k,

then (a−b)  is also divisible by 3.

Case(2)  (a−b) but not (a+b)  is divisible by 3

If a is chosen from (3k+1) set ,then b must be from (3k+1) set .

Same for (3k+2) set.

So, number of ways = 2* 10*10

Number of ways of choosing such that a,b such that (a  

2

−b  

2

) is divisible by 3

=30*30

Probability  

=  

30×30

5×10×10

​  

 

=  

9

5

​

Step-by-step explanation:

hope u ike