# The product of two consecutive odd integers is 783. How do you find the integers?

Jan 2, 2017

Here's how you can do that.

#### Explanation:

The problem tells you that the product of two consecutive odd integers is equal to $783$.

Right from the start, you know that you can get from the smaller number to the bigger number by adding $2$.

You need to add $2$ because if you start with an odd number and add $1$, you end up with an even number, which is not supposed to happen here.

$\text{odd number" + 1 = "the consecutive even number"" } \textcolor{red}{\times}$

$\text{odd number" + 2 = "the consecutive odd number"" } \textcolor{\mathrm{da} r k g r e e n}{\sqrt{}}$

So, if you take $x$ to be the first number, you can say that

$x + 2$

is the second number, which means that you have

$x \cdot \left(x + 2\right) = 783$

$\frac{\textcolor{w h i t e}{a}}{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a}}$
SIDE NOTE You can also go with $x - 2$ as the first number and

$\left(x - 2\right) + 2 = x$

as the second number, the answer must come out the same.
$\frac{\textcolor{w h i t e}{a}}{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a}}$

This is equivalent to

${x}^{2} + 2 x = 783$

${x}^{2} + 2 x - 783 = 0$

Use the quadratic formula to find the two values of $x$ that satisfy this equation

${x}_{1 , 2} = \frac{- 2 \pm \sqrt{{2}^{2} - 4 \cdot 1 \cdot \left(- 783\right)}}{2 \cdot 1}$

${x}_{1 , 2} = \frac{- 2 \pm \sqrt{3136}}{2}$

${x}_{1 , 2} = \frac{- 2 \pm 56}{2} \implies \left\{\begin{matrix}{x}_{1} = \frac{- 2 - 56}{2} = - 29 \\ {x}_{2} = \frac{- 2 + 56}{2} = 27\end{matrix}\right.$

Now, you have two valid solution sets here.

• $\text{For} \textcolor{w h i t e}{.} x = - 29$

$- 29 \text{ }$ and $\text{ } - 29 + 2 = - 27$

Check:

$\left(- 29\right) \cdot \left(- 27\right) = 783 \text{ } \textcolor{\mathrm{da} r k g r e e n}{\sqrt{}}$

• $\text{For} \textcolor{w h i t e}{.} x = 27$

$27 \text{ }$ and $\text{ } 27 + 2 = 29$

Check:

$27 \cdot 29 = 783 \text{ } \textcolor{\mathrm{da} r k g r e e n}{\sqrt{}}$

Jan 2, 2017

There are two solutions:

$27 , 29$

and

$- 29 , - 27$

#### Explanation:

One method goes as follows.

I will use the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

Let $n$ denote the even number between the consecutive odd integers $n - 1$ and $n + 1$.

Then:

$783 = \left(n - 1\right) \left(n + 1\right) = {n}^{2} - 1$

Subtract $783$ from both sides to get:

$0 = {n}^{2} - 784 = {n}^{2} - {28}^{2} = \left(n - 28\right) \left(n + 28\right)$

So $n = \pm 28$

There are therefore two possible pairs of consecutive odd integers:

$27 , 29$

and:

$- 29 , - 27$

Jan 2, 2017

Find $\sqrt{783}$

$27 \times 29 = 783 \text{ and } - 27 \times - 29 = 783$

#### Explanation:

We know from the question that $783$ is the product of 2 numbers, which means they are factors.

We also know that the two factors are very close together because they are consecutive odd numbers.

If you consider factor pairs you will find that the closer factors are, the smaller is their sum or difference.

The factors which are furthest apart are $1 \mathmr{and} 783$

The factors which have the smallest sum or difference are the square roots. The square root of a number is the factor exactly in the middle if factors arranged in order.

$1 \text{ "3" "9 ...... sqrt783 ......87" "261" } 783$

The factors we are looking for must be very close to $\sqrt{783}$

$\sqrt{783} = 27.982 \ldots . .$

Test odd numbers on either side of $27.982 \ldots$

$27 \times 29 = 783 \text{ } \leftarrow$ and VOILA!!

Remember that the odd numbers can be negative as well.