# Three consecutive odd numbers have a sum of 45. What is the greatest number?

Nov 29, 2016

The greatest odd number is $17$

#### Explanation:

Let the three consecutive odd numbers are $2 n + 1 , 2 n + 3 \mathmr{and} 2 n + 5$

Their sum is $\left(2 n + 1\right) + \left(2 n + 3\right) + \left(2 n + 5\right) = 45 \mathmr{and} 6 n + 9 = 45 \mathmr{and} 6 n = 36 \mathmr{and} n = 6$

The greatest odd number is $2 n + 5 = 2 \cdot 6 + 5 = 17$[Ans]

Nov 29, 2016

Slightly different start point but gives the same answer.

17

#### Explanation:

You do not need to define a sequence that guarantees you are looking at odd numbers.

However, you must guarantees they are all of same type. That is all odd or all even.

You do this as follows

Let the first number be $n$
Then the next number of same type (odd or even) is $n + 2$
The number after that is $n + 4$

So we have:

$n + \left(n + 2\right) + \left(n + 4\right) = 45$

$3 n + 6 = 45$

$3 n = 39$

$n = \frac{39}{3} = 13$

So the greatest number is $n + 4 \text{ "=" "13+4" "=" } 17$