# The product of two consecutive negative integers is 1122. How do you find the integers?

Jul 22, 2017

$- 34 , \mathmr{and} , - 33.$

#### Explanation:

As we need two consecutive integers, let us suppose that, these

are $x \mathmr{and} x + 1 , \text{ where, } x \mathmr{and} \left(x + 1\right) < 0.$

By what is given, $x \left(x + 1\right) = 1122.$

$\therefore {x}^{2} + x = 1122.$

Multiplying by $4 , \text{ we get, } 4 {x}^{2} + 4 x = 4488.$

Completing the square, we have,

$4 {x}^{2} + 4 x + 1 = 4489.$

${\left(2 x + 1\right)}^{2} = {67}^{2.}$

$2 x + 1 = \pm 67.$

because x < 0, &, (x+1) < 0; :. x+(x+1)=2x+1 < 0.

$\therefore 2 x + 1 = - 67 \Rightarrow 2 x = - 67 - 1 = - 68.$

$\therefore x = - 34 , \mathmr{and} , x + 1 = - 33.$

Jul 22, 2017

Set up the equation $n \times \left(n + 1\right) = 1122$ solve for n and then find n+ 1

$- 33 \times - 34 = 1122$

#### Explanation:

$n \times \left(n + 1\right) = 1122$ This gives

${n}^{2} + 1 n = 1122$ subtracting both sides by 1122 gives

${n}^{2} + 1 n - 1122 = 1122 - 1122$ or

${n}^{2} + 1 n - 1122 = 0$

$1122 = 33 \times 34$

$\left(n - 33\right) \times \left(n + 34\right) = 0$

$n = 33 \mathmr{and} n = - 34$

Negative factors are asked for, so reject $n = 33$

The factors are $- 33 \mathmr{and} - 34$

Jul 22, 2017

The two integers are $- \left\lceil \sqrt{1122} \right\rceil = - 34$ and $- \left\lfloor \sqrt{1122} \right\rfloor = - 33$

#### Explanation:

If $0 < a < b$ and $a b = n$ then:

${a}^{2} < a b = n < {b}^{2}$

where all parts are positive. So taking the square root, we find:

$a < \sqrt{n} < b$

So if $a$ and $b$ are consecutive positive integers then:

$a = \left\lfloor \sqrt{n} \right\rfloor \text{ }$ and $\text{ } b = \left\lceil \sqrt{n} \right\rceil$

For $n = 1122$, we find $\sqrt{n} \approx 33.496$

So:

$\left\lfloor \sqrt{n} \right\rfloor = 33$

and

$\left\lceil \sqrt{n} \right\rceil = 34$

We find:

$33 \cdot 34 = 1122$

as required.

Hence the negative solutions are $- 34$ and $- 33$ since:

$\left(- 34\right) \left(- 33\right) = 34 \cdot 33 = 1122$

$\textcolor{w h i t e}{}$
Hmmm - OK but how do you find $\sqrt{1122}$?

All we need is a reasonable approximation for our purposes.

${30}^{2} = 900 < 1122 < 1600 = {40}^{2}$

So:

$30 < \sqrt{1122} < 40$

Next, linearly interpolate:

$\sqrt{1122} \approx 30 + \frac{1122 - 900}{1600 - 900} \cdot 10 = 30 + \frac{222}{700} \cdot 10 \approx 33$

Then we can use:

$\sqrt{{a}^{2} + b} = a + \frac{b}{2 a + \frac{b}{2 a + \frac{b}{2 a + \ldots}}}$

with $a = 33$ and $b = 1122 - {33}^{2} = 1122 - 1089 = 33$

So:

$\sqrt{1122} = 33 + \frac{33}{66 + \frac{33}{66 + \frac{33}{66 + \ldots}}}$

which we can quickly see is about $33 \frac{1}{2}$, which is good enough for our purposes.

Or we could have simply noticed: Oh look, $1122 - {33}^{2} = 33$ so $1122 = 33 \cdot 34$, which is what we wanted to know.

Jul 22, 2017

$- 34 \mathmr{and} - 33$

#### Explanation:

Note what happens when the factors of a number are arranged in order:

For example, $36$

$1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36$

The 'outer pair' $1 \mathmr{and} 36$:
have the greatest sum $1 + 36 = 37$
and the greatest difference: $36 - 1 = 35$

The 'pair' in the middle would be $6 \mathmr{and} 6$, but we only write one $6$
This gives the smallest sum, $6 + 6 = 12$
and the smallest difference: $6 - 6 = 0$

Also note that there is a single factor exactly in the middle because $36$ is a perfect square and $\sqrt{36} = 6$

A number such as $12$ does not have a square root, but the middle factors are consecutive numbers, $3 \mathmr{and} 4$

$12 : 1 , 2 , \textcolor{b l u e}{3 , 4} , 6 , 12$
$\textcolor{w h i t e}{\times \times \times} \uparrow$
$\textcolor{w h i t e}{\times \times \times} \sqrt{12}$

$\sqrt{12}$ lies between $3 \mathmr{and} 4$

Therefore the same will happen with consecutive factors of $1122$, they will lie on either side of $\sqrt{1122}$

$\sqrt{1122} = 33.496$

The negative integers on either side are $- 34 \mathmr{and} - 33$

Check: $- 34 \times - 33 = 1122$