Let A=1,a_1,a_2,,a_18,77 be a set of integers with 1

Question

Let $A=\{1,a_1,a_2,\ldots,a_{18},77\}$ be a set of integers with $1<a_1<a_2<\cdots<a_{18}<77$. Let $A+A=\{x+y:x,y\in A\}$ contain exactly 39 elements. Then the value of $a_1+a_2+\cdots+a_{18}$ is equal to\_\_\_

Answer: 702

Accepted Answer

For $A+A$ to have the minimum possible number of elements ($2\times20-1=39$), $A$ must be an arithmetic progression. With first term 1 and last term 77, and 20 terms: common difference $=\dfrac{77-1}{19}=4$.

So $A = 1,5,9,13,\ldots,77$, meaning $a_1,\ldots,a_{18} = 5,9,13,\ldots,73$ (18 terms, an A.P. with first term 5, common difference 4).

$$\text{Sum} = \frac{18}{2}(5+73) = 9\times78 = 702$$

Solutions · 1