A ten digit number is taken in such a way that it is lying between 9582610000 and 9582620000. The last 4 digit of the number are 4 successive digits not necessarily taken in order. Average of these 4 digits is 5.5. On forming a new 4 digit number by taking these 4 digits in reverse order, we find a 4 digit number divisible by 2 but not by 4. The 10 digit number is divisible by 25. hundreds place of the number will be
OPtion
1) 6
2) 8
3) 2
4) 7
5) 9
6) 4
7) 1
8) 3
9) 0
10)5
Solution
Let number is x
Given that
9582610000 < x < 9582620000
Sum of last 4 digit = 5.5*4 = 22
=> n + n+1 + n+2 + n+3 = 22
=> n = 4
So digit are 4,5,6,7 but not necessarily in order
10 digit number is divisible by 25 => last 2 digit are 75 in order so possible cases are
6475 and 4675
On making a number by reversing the digits we get 5746 or 5764. Both are divisible by 2 but 5746 is not divisible by 4 so last 4 digits are 6475 in order so number will be 9582616475
Hundreds place of the number will be 4