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.
Select the correct option
A) 2
B) 3
C) 4
D) 5
E) 6
Let us understand the meaning of |a| &gt; 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)| &lt; 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)| &gt; 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| &gt; 2 ==&gt; a &gt; 2 i.e. a can have any value bigger than 2 (i.e. 3, 4, 5…).   Now, lets move to equation a + 2b = 32 ==&gt; b = (32 – a)/2 ==&gt; 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 4 (value 4).
<result>C</result>
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.