Permutations & Combinations
Sumset with exactly 39 elements; AP constraint
MMTS_Full_Test_01
Grade 12
Question:
Let $A = \{1, a_1, a_2, \ldots, a_{18}, 77\}$ be a set of integers with $1 < a_1 < a_2 < \ldots < a_{18} < 77$. Let $A+A = \{x+y : y\in A\}$ contain exactly 39 elements. Then $a_1+a_2+\ldots+a_{18}$ equals
(A) 600
(B) 702
(C) 800
(D) 200
Step-by-Step Solution
Key Concept: For $|A+A|$ to be minimal ($=2|A|-1=39$), $A$ must be an arithmetic progression. Determine the common difference.
$d=4$, AP: $1,5,9,\ldots,77$. $\sum a_i=\frac{18}{2}[2(5)+17(4)]=702$.
Correct Answer: (B) 702