How do you find two consecutive positive odd integers whose product is 143?

Feb 6, 2016

$143 = 11 \times 13$

Explanation:

Notice that consecutive odd integers will differ by $2$, so we could call the even number between them $n$, making the two integers $n - 1$ and $n + 1$.

So:

$143 = \left(n - 1\right) \left(n + 1\right) = {n}^{2} - 1$

Add $1$ to both ends and transpose to get:

${n}^{2} = 144 = {12}^{2}$

So $n = \pm 12$

Since we are told that the integers are positive, use $n = 12$ to find the pair of integers: $11 , 13$.

Alternatively, just find factors of $143$, trying prime numbers in turn: $2$ (no), $3$ (no), $5$ (no), $7$ (no), $11$ (yes). Then $\frac{143}{11} = 13$ - we have found our pair of numbers.

Feb 6, 2016

11 and 13

Explanation:

Proceed as below.
Let $\left(2 n + 1\right)$ be the first odd integer, where $n$ is any positive integer. Second number being consecutive is $\left(2 n + 3\right)$
Rest of the steps are same as followed by others.

It is by convention, as $2 n + 1$ is always odd.

Feb 6, 2016

Expanding on AO8's answer: This approach guarantees that any number we look at is odd:$\text{ }$Solution: 11 & 13

Explanation:

We need to be able to guarantee that we are dealing with odd numbers.

Let $n$ be any number where $n > 0$

Suppose n is even. Then $2 n$ is also even.
Suppose n is odd. Then $2 n$ is also even.

But in both cases $2 n + 1$ will be odd.

Let the first odd number be $2 n + 1$
Then the second odd number will be $2 n + 3$

We are given that the product is 143 so write the equation as:

$\left(2 n + 1\right) \left(2 n + 3\right) = 143$

This is saying exactly the same thing as the others but in a slightly different way! So we now have:

$4 {n}^{2} + 8 n - 140 = 0$

Divide by 4

${n}^{2} + 2 n - 35 \text{ "=" "(n+7)(n-5)" "=" } 0$

So for my conditions $n = - 7 \text{ or } + 5$

$n = - 7 \text{ does not work so we use } n = 5$

Thus the first number is: $\left(2 n + 1\right) \text{ "->" } 2 \left(5\right) + 1 = 11$

So the second number is $11 + 2 = 13$