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

##### 4 Answers

#### Explanation:

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

are

By what is given,

Multiplying by

Completing the square, we have,

Set up the equation

#### Explanation:

Negative factors are asked for, so reject

The factors are

The two integers are

#### Explanation:

If

#a^2 < ab=n < b^2#

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

#a < sqrt(n) < b#

So if

#a = floor(sqrt(n))" "# and#" "b = ceil(sqrt(n))#

For

So:

#floor(sqrt(n)) = 33#

and

#ceil(sqrt(n)) = 34#

We find:

#33*34 = 1122#

as required.

Hence the negative solutions are

#(-34)(-33) = 34*33 = 1122#

**Hmmm - OK but how do you find #sqrt(1122)#?**

All we need is a reasonable approximation for our purposes.

Start with:

#30^2 = 900 < 1122 < 1600 = 40^2#

So:

#30 < sqrt(1122) < 40#

Next, linearly interpolate:

#sqrt(1122) ~~ 30+(1122-900)/(1600-900)*10 = 30+222/700*10 ~~ 33#

Then we can use:

#sqrt(a^2+b) = a+b/(2a+b/(2a+b/(2a+...)))#

with

So:

#sqrt(1122) = 33+33/(66+33/(66+33/(66+...)))#

which we can quickly see is about

Or we could have simply noticed: Oh look,

#### Explanation:

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

For example,

The 'outer pair'

have the greatest sum

and the greatest difference:

The 'pair' in the middle would be

This gives the smallest sum,

and the smallest difference:

Also note that there is a single factor exactly in the middle because

A number such as

Therefore the same will happen with consecutive factors of

The negative integers on either side are

Check: