JAMB Mathematics 2011 Paper

Concept:Four consecutive integers differ by $1$.If the smallest integer is $n$, then the others are $n+1$, $n+2$, and $n+3$.Their sum equals $34$.Explanation:Write the sum of the four integers:$$n + (n+1) + (n+2) + (n+3) = 34$$Combine the $n$ terms and the constant terms:$$4n + 6 = 34$$Subtract $6$ from both sides:$$4n = 28$$Divide both sides by $4$:$$n = \frac{28}{4} = 7$$So the integers are $7$, $8$, $9$, and $10$.Their sum is $7 + 8 + 9 + 10 = 34$, which confirms the answer.The least of these four numbers is $7$.Answer:The least number is $7$ (Option A).