Three positive integers a, b, and c are such that their average is 40 and a ≤ b ≤ c. If the median is (a + 13), what is the least possible value of c?  Answers:	A) 41	B) 44	C) 45	D) 46	E) 47
Solution   Given Data   3 positive integers   Average of a, b, and c = 40.   So, a+b+c3a+b+c3 =120 or a + b + c = 120   a ≤ b ≤ c   Median = (a + 13)   Median of these 3 numbers a, b, and c is ‘b’ because a ≤ b ≤ c   Therefore, b = a + 13   Objective To find the least value of 'c'   Theoretically, the least value of c is when c = b.   Therefore, a + (a + 13) + (a + 13) = 120 (b and c are equal and b, the median, is a + 13)   Or 3a =94 or a = 31.33   So, b = c = 31.33 + 13= 
<gadget id="calculator">31.33 + 13</gadget>
<output>44.33</output>
44.33   However, we know that these numbers are all integers.   Therefore, a, b, and c cannot take these values.   So, the least value for c with this constraint is NOT likely to be when c= b   Let us increment c by 1. Let c = (b + 1)   In this scenario, a + (a + 13) + (a + 14) = 120   Or 3a = 93 or a=31 The value of the numbers is an integer   If a = 31, b = 44 and c = 45.   The least value for c that satisfies all these conditions is 45.   45orrect answer is choice 45.
<result>C</result>
Three positive integers a, b, and c are such that their average is 40 and a ≤ b ≤ c. If the median is (a + 13), what is the least possible value of c?