# The sum of two consecutive integers is 68, what is the smaller number?

Jul 18, 2016

$\textcolor{red}{\text{This question is wrong!}}$

#### Explanation:

$\textcolor{b l u e}{\text{Why this question is wrong}}$

Two consecutive number means that one of them is even and the other odd. Consequently their sum will be odd.

For the sum to be 68 the question has to be one of:
Two consecutive even numbers give an even number answer.
Two consecutive odd numbers give an even number answer.
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$\textcolor{b r o w n}{\text{Alternative questions}}$

$\textcolor{b l u e}{\text{Solution for two consecutive even numbers sum to 68}}$

Let $n$ be any number

Then $2 n$ is even

So $2 n + 2$ is the next even number

Thus $2 n + \left(2 n + 2\right) = 68$

So $4 n + 2 = 68$

Subtract 2 from both sides

$4 n = 66$

$n = \frac{66}{4} = 16.5 \leftarrow \text{ seed value}$

Thus the 1st even number is $2 n \to 2 \times 16.5 = 33$
Thus the next even number is $33 + 2 = 35$

$\textcolor{b l u e}{33 + 35 = 68}$
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$\textcolor{b l u e}{\text{Solution for two consecutive odd numbers sum to 68}}$

Using the notation from the first solution

If $2 n$ is even then $2 n + 1$ is odd and the first number
The second odd number will be $\left(2 n + 1\right) + 2 = 2 n + 3$

So $\left(2 n + 1\right) + \left(2 n + 3\right) = 68$

$\implies 4 n + 4 = 68$

$\implies 4 n = 64$

Divide both sides by 4

$\implies n = \frac{64}{4} = 16 \leftarrow \text{ Seed value}$

So the first odd number is $2 n + 1 = 2 \left(16\right) + 1 = 33$
So the second odd number is $33 + 2 = 35$

$\textcolor{b l u e}{33 + 35 = 68}$