A Quick Way to Solve 1000^2 − 999^2: The Difference of Two Squares
Speeding Up Your Arithmetic
Suppose you needed to do the sum 1000^2 − 999^2. What would you do? Pull out a calculator? Panic? No need. While writing this sentence, I quickly calculated in my head that the answer is 1999.
But how did I manage that so quickly? I'm not a human calculator, and I don't know all of the squares off the top of my head. Instead, I used a very simple, quick trick to solve this seemingly difficult sum. This trick is called the difference of two squares.
But how did I manage that so quickly? I'm not a human calculator, and I don't know all of the squares off the top of my head. Instead, I used a very simple, quick trick to solve this seemingly difficult sum. This trick is called the difference of two squares.
What Is the Difference of Two Squares Trick?
First of all, let's look at this algebraically. Our sum is of the form a^2 - b^2, where a and b are both integers (whole numbers). An interesting property of a2 - b2 is that it can be factorised into two brackets like this:
a^2 − b^2 = a^2 − ab + ab − b^2
= a(a − b) + b(a − b)
= (a + b)(a − b)
Using this, we can quickly speed up our sum.
1000^2 − 999^2 = (1000 + 999)(1000 − 999)
And there we have it. A seemingly difficult calculation is turned into something very simple and easy to do mentally.
= 1999 × 1
= 1999
a^2 − b^2 = a^2 − ab + ab − b^2
= a(a − b) + b(a − b)
= (a + b)(a − b)
Using this, we can quickly speed up our sum.
1000^2 − 999^2 = (1000 + 999)(1000 − 999)
And there we have it. A seemingly difficult calculation is turned into something very simple and easy to do mentally.
= 1999 × 1
= 1999
Generalising Our Result
We can use this trick to quickly solve any subtraction using consecutive squares.
If we take a perfect square, n2, and the next perfect square, (n+1)^2 for some integer n, then:
(n+1)^2 − n^2 = ((n + 1) + n) ((n + 1) − n)
= (2n + 1) × 1
= 2n + 1
So, using this result, we can calculate that:
30^2 − 29^2 = 2 × 29 + 1
= 59
or
950^2 − 949^2 = 2 × 949 + 1
= 1899
and so on for any consecutive perfect squares that we like.
If we take a perfect square, n2, and the next perfect square, (n+1)^2 for some integer n, then:
(n+1)^2 − n^2 = ((n + 1) + n) ((n + 1) − n)
= (2n + 1) × 1
= 2n + 1
So, using this result, we can calculate that:
30^2 − 29^2 = 2 × 29 + 1
= 59
or
950^2 − 949^2 = 2 × 949 + 1
= 1899
and so on for any consecutive perfect squares that we like.
Another Use of the Difference of Two Squares for Quick Arithmetic
We can also use the difference of two squares to quickly complete the multiplication of two numbers (as long as their average can be easily squared).
For example, take 38 × 42. The average of these two numbers is 40, which can be easily squared to 1600. We can then use the difference of two squares as such:
38 × 42 = (40 − 2)(40 + 2)
= 40^2 − 2^2
= 1600 − 4
= 1596
We can also use it to quickly calculate 105 × 95:
105 × 95 = (100 + 5)(100 − 5)
= 100^2 − 5^2
= 10000 − 25
= 9975
For example, take 38 × 42. The average of these two numbers is 40, which can be easily squared to 1600. We can then use the difference of two squares as such:
38 × 42 = (40 − 2)(40 + 2)
= 40^2 − 2^2
= 1600 − 4
= 1596
We can also use it to quickly calculate 105 × 95:
105 × 95 = (100 + 5)(100 − 5)
= 100^2 − 5^2
= 10000 − 25
= 9975
Comments
Have you ever used this trick?
Tell us about it in the comments below.
Tell us about it in the comments below.