JEE Main 26 June 2022 Shift 2 Solved Paper

Total number of numbers from given
Condition =n(s)=26
Every required number is of the form
A=7.(10a1+10a2+10a3+......)+111111
Here 111111 is always divisible by 21.
∴ If A is divisible by 21 then
10a1+10a2+10a3+...... must be divisible by 3.
For this we have ‌6C0+‌6C3+‌6C6 cases are there
∴n(E)=‌6C0+‌6C3+‌6C6=22
∴ Required probability =‌

22

26

=p
∴‌

11

32

=p
∴96p=33