32 = a + 2b
|a| > 2
If ‘a’ and ‘b’ are both integers, then what is the smallest possible values of ‘a’ that can be used to solve the above equation. Options:  A) 2  B) 3  C) 4  D) 5  E) 6

Rationale: let us understand the meaning of | a | > 2 mod is very easy concept if you solve mod question by considering as a distance . when a mod is written as | x - ( a ) | = b , this means the distance from point ' a ' ( both side left and right of ' a ' on number line ) is b . | x - ( a ) | < b means the distance is between the two extreme distance ( left and right side of ' a ' on number line , considering the max distance is ' b ' from ' a ' - as per this scenario . . . . . hence the value of ' a ' must be between these two extremes . | x - ( a ) | > b means the distance is greater than the distance of ' b ' . . i . e the value of a could be anywhere more than ' b ' ( i . e . current case ) . now come to the question . first its given | a | > 2 = = > a > 2 i . e . a can have any value bigger than 2 ( i . e . 3 , 4 , 5 … ) . now , lets move to equation a + 2 b = 32 = = > b = ( 32 – a ) / 2 = = > b = 16 – ( a / 2 ) . according to question , b is an integer , hence to make ‘ b ’ integer a must be divisible by 2 . minimum possible value of ‘ a ’ that is divisible by 2 is 4 ( a is greater than 2 so the next number that is divisible by 2 is 4 ) . so the answer is c ( value 4 ) .
Answer: C