If the sum of three consecutive odd integers is p, what is the greatest of the three integers?

Question:

If the sum of three consecutive odd integers is p, what is the greatest of the three integers?

Odd Numbers:

(i) An odd number is a number that leaves a remainder of one when divided by two. For example, the odd numbers are {eq}1,3,5,... {/eq}.

(ii) The difference between two consecutive odd numbers is always two. So we can assume the consecutive odd numbers to be {eq}x, x+2, x+4, ... {/eq}.

Let us assume the three consecutive odd integers are {eq}x, x+2 \text{ and }x+4 {/eq}.

Their sum is given to be {eq}p {/eq}.

So we get the equation:

$$x+(x+2)+(x+4)=p \\[0.3cm] 3x+ 6 = p \\[0.3cm] \text{Subtracting 6 from both sides},\\[0.3cm] 3x = p-6\\[0.3cm] \text{Dividing both sides by 3},\\[0.3cm] x= \dfrac{p-6}{3}$$

The greatest number among {eq}x, x+2 \text{ and }x+4 {/eq} is {eq}x+4 {/eq}.

Substitute {eq}x= \dfrac{p-6}{3} {/eq} in the above number:

\begin{align} x+4 &=\dfrac{p-6}{3} + 4 \\[0.3cm] &= \dfrac{p-6}{3} + \dfrac{12}{3}\\[0.3cm] &= \dfrac{p-6+12}{3}\\[0.3cm] &= \color{blue}{\boxed{\mathbf{\dfrac{p+6}{3}}}} \end{align}