Answer to Question #125712 in Math for Nomzamo

Question

Lester selected the following pattern of numbers (6 ,12,13,14,20) from the calendar. He says that anywhere in the calendar, for numbers arranged in this pattern, the sum of the outside four numbers will be four times the number in the middle. Do you think that Lester is correct? How can you check? Please explain.

Answer 1

$$\begin{matrix} & (n) & \\ (n+7-1) & \colorbox{yellow}{(n+7)} & (n+7+1) \\ & (n+7+7) & \end{matrix}$$

$$n+(n+7-1)+(n+7+1)+(n+7+7)=4n+28$$ $$=4n+28=4(n+7)$$

Anywhere in the calendar, for numbers arranged in this pattern, the sum of the outside four numbers will be four times the number in the middle. Lester is correct.