# The product of two consecutive odd integers is 143. How do you find the integers?

Dec 25, 2016

11 and 13.

#### Explanation:

Call x a number
The number (2x + 1) should be an odd number no matter what x is.
The consecutive odd number to (2x + 1) will be: (2x + 3).
There for:
(2x + 1)(2x + 3) = 143
$4 {x}^{2} + 6 x + 2 x + 3 = 143$
$4 {x}^{2} + 8 x - 140 = 0$
${x}^{2} + 2 x - 35 = 0$
Solve this quadratic equation for x. Find 2 numbers knowing sum (-2) and product (-35). They are x = 5 and x = -7 (rejected as negative)
The odd number is N = (2x + 1) = 10 + 1 = 11
The consecutive odd number is: N + 2 = (2x + 3) = 10 + 3 = 13
Their product is (11 x 13) = 143. OK

Dec 25, 2016

Two solutions:

$11$ and $13$

$- 13$ and $- 11$

#### Explanation:

Let $n$ be the even number between the two consecutive odd numbers $n - 1$ and $n + 1$.

Then we have:

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

Add $1$ to both ends to get:

$144 = {n}^{2}$

So $n = \pm \sqrt{144} = \pm 12$

If $n = 12$ then the two odd numbers are $11$ and $13$.

If $n = - 12$ then the two odd numbers are $- 13$ and $- 11$.

Dec 25, 2016

Find $\sqrt{143}$, then use trial and error

$11 \times 13 = 143 \text{ and } - 11 \times - 13 = 143$

#### Explanation:

The consecutive odd numbers must lie on either side of the square root of 143

$\sqrt{144} = 12$ so $\sqrt{143} \approx 11.9$

Look for odd numbers either side of $\sqrt{143}$.

Try $11 \mathmr{and} 13$

$11 \times 13 = 143 \text{ } \leftarrow$ these are the factors we need.

Remember the factors could be negative as well..

$- 11 \times - 13 = 143$

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Background knowledge:
If you write the factors of a number in increasing order, note the following:

Factors are always in pairs.

• The difference between the factors is greatest for the outside factors and least for the innermost factors.

For example the factors of 48 are:

$1 \text{ "2" "3" "4" "6" "8" "12" "16" "24" } 48$

48- 1 = 47 but 8-6 = 2

• The square root of a number is exactly in the middle of the factors

For example the factors of 36 are

$1 \text{ "2" "3" "4" "6" "9" "12" "18" } 36$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .} \uparrow$
#color(white)(.......................)sqrt36

$6 - 6 = 0$ this is the smallest possible difference.