24 = 3a + 5b
3 > |–a|
If ‘a’ and ‘b’ are both integers, then what is the highest possible values of ‘b’ that can be used to solve the above equation.  Choose the correct choice from the following answers:
A) 2	B) 3	C) 4	D) 5	E) 6
Let us understand the meaning of 5 &gt; |-a|   1) when we write |-x| this means the same as |x|.   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'.   Now come to the question. First its given|-a| &lt; 3 ==&gt; |a| &lt; 3 ===&gt; |a-0| &lt; 3 ==&gt; the distance from zero is less than 3. So the point ‘a’ will be -2,-1,0,1,2 as distance from 0 to these values is less than 3.   Now, lets move to equation 3a + 5b = 24 ==&gt; b = (24 – (3 * a))/5. According to question, b is an integer, hence to make ‘b’ integer (24 – (3*a)) must be divisible by 5. Now remove the value which can’t make (24-(3*a)) divisible by 5 from the possible values of a. It will remain is -2. With only one value of ‘a’ we can have only one value of ‘b’ and that is b = (24 – 3 *(-2))/5 ==&gt; b = (24 + 6)/5 ==&gt; b = 6. Hence the highest value that b can take is 6 and the answer is 6 (value 6).
<result>E</result>
24 = 3a + 5b
3 > |–a|
If ‘a’ and ‘b’ are both integers, then what is the highest possible values of ‘b’ that can be used to solve the above equation.