Question

let a,b,c,d be non zero distinct digit. the number of 4 digit numbers abcd such that ab+cd is even is divisible by. option - 3. , 4 , 7 , 11 .

Answer 1

So firstly deduce the conditions a little bit
What I mean is that
ab+cd= even
which is only possible
when ab=odd and cd=even
or
ab=even and cd=even
what I further suggest is that now since it is a mcq assume the values of ab and cd which satisfies the above condition
Note:While assumption don't assume integers like 0 and 1 as they are risky they sometimes satisfy only special results
So now I m assuming 4 non zero distinct digits
Taking all even would be a easier choice
Let's assume a=2 b=4 c=6 d=8
now
=ab+cd
=2*4+6*8
=56 which is divisible by both 7 and 4
so now try another set of values
ab+cd
=2*4+6*10
=68
which is only divisible by 4
Therefore the answer b
The answer seems to be lengthy
but I while doing u can do this even orally
HOPE THIS HELPS
MARK AS BRAINLIEST

Answer 2

7392 the required answer.
LCM of 3, 4, 7, 11 is 7392
7392 is also an even number

7392 / 3 = 2464
7392 / 4 = 1848
7392 / 7 = 1056
7392 / 11 = 672