Question

If the sum of the last $3$ integers in a set of $6$ consecutive integers is $624$, what is the sum of the first $3$ integers of the set?

• A. $515$
• B. $600$
• C. $615$
• D. $750$

Answer 1

Answer: (C)

$615$

Think of the set of integers as $n,(n+1),(n+2),(n+3),(n+4)$, and $(n+5)$. Thus, $(n+3)+(n+4)+(n+5)=3n+12=624$. Therefore, $n=204$.

The sum of the first three integers is: $204+205+206=615$.

Alternatively, another way you could solve this algebraically is to line up the algebraic expressions for each number so that you can subtract one from the other directly:

Sum of the last three integers. $(n+3)+(n+4)+(n+5)$

Less the sum of the first three integers. $\underset{–}{-[n+(n+1)+(n+2\left)\right]}$

$3+3+3=9$

Thus, the sum of the last three numbers is $9$ greater than the sum of the first three numbers, so the sum of the first three numbers is $624-9=615$.

Visually, you can represent the six consecutive unknowns with six lines:

$Sum=Average\times 3$

$=205\times 3=615$.