The sum of three consecutive odd integers is 48, how do you find the largest integer?

Apr 25, 2017

The question has the wrong value as the sum. Summing 3 odd numbers will give an odd sum. However; the method is demonstrated through an example

Explanation:

Just to make this work lets derive the sum first. Suppose we had

$9 + 11 + 13 = 33$ as our initial odd number

Let the fist odd number be $n$

Then the second odd number is $n + 2$

Then the third odd number is $n + 4$

So we have:

$n + \left(n + 2\right) + \left(n + 4\right) = 33$

$3 n + 6 = 33$

Subtract 6 from both sides

$3 n = 27$

Divide both sides by 3

$n = 9$

So the largest number is $9 + 4 = 13$

Apr 25, 2017

Explanation below.

Explanation:

The question is worded incorrectly because there are not three consecutive odd integers that add up to $48$.

What I can do for you is leave you with this method of solving this problem. Let's say I was looking for 3 consecutive integers that add up to $81$.

My first integer would be $2 x - 1$
My second integer would be $2 x + 1$
My third integer would be $2 x + 3$

So my equation is...

$2 x - 1 + 2 x + 1 + 2 x + 3 = 81$

$6 x + 3 = 81$

$6 x = 81 - 3$

$6 x = 78$

$\cancel{6} \frac{x}{\cancel{6}} = \frac{78}{6}$

$x = 13$

Now we know the value of $x$ so we plug it into our 3 equations.

My first integer would be $2 \left(13\right) - 1$ $- - \to$ $= 25$
My second integer would be $2 \left(13\right) + 1$$- - \to$ $= 27$
My third integer would be $2 \left(13\right) + 3$$- - \to$ $= 29$

So,

$25 + 27 + 29 = 81$