If a and b are positive integers, is the sum a + b divisible by 4?

So we require to know the value of a+b or the last two digits as divisibility by 4 depends on last two digits.

(1) \(23^a+25^b\) divided by 10 gives remainder 8.
This means that the last digits of a and b add up to 8.
25^b will leave a remainder 5, whatever be the value of b.
To get the sum as 8, we should get the last digit of 23^a as 3. The cycilicty of 3 is 3,9,7,1,3,9,7,1..., so a can be 1,5,9.. or a has to be of type 4K+1.
If b is 3,7... or of type 4x-1, answer is yes, otherwise no.
Insuff

(2) 22^b divided by 10 gives a remainder 8.
Cyclicity of 2 is 2,4,8,6,2,4,8,6.... But we are looking for last digits as 8, so b can be 3,7.....or of type 4b-1.
Nothing known about a.
Insuff

Combined
a is of type 4K+1 and b is of type 4b-1, so a+b=4K+1+4b-1=4(k+b), which is divisible by 4.
Sufficient


C

a+b sum divisible by 4
#1
When the sum 23^a+25^b is divided by 10, the remainder is 8
possible when a is odd integer values possible 1,5,9,13
and b is odd integer value ; 1,3,5,7,9..
we get yes & no a+b ; 1+1 ; no 5+3 ; yes
insufficient
#2
When 22^b is divided by 10, the remainder is 8
possible with values of b being 3,7,11,15
value of a not know insufficient
from 1 &2
when b is 3 a would be 5 i.e sum of both is divisible by 4
sufficient
option C