Multiplication with Squares

Multiplication using the nearest square

When 2 numbers being multiplied are not too far apart this method is pretty efficient
if the squares table above is remembered.
This method is simple.
v x ( v + d ) = product
v is the average of the two numbers ( truncated)
If the 2 numbers are a & b then
v = (a+b)/2 ( decimals are dropped )
If a > b then d = a - v

If a + b = even then product = v x v - d x d
otherwise product = v x v + v - d x d + d

To use this method, follow the following procedure :

step 1 : find v & v²
step 2: determine if a+b is even or odd
this is known during step 1 when you do a + b
step 3: d = a - v
step 3a ( a+b is even) calculate v² - dxd
step 3b ( a+b is odd ) calculate v² + v - dxd + d

Strategy - Remembering the squares

If n² is a known square, then :

the square of n + 1 = n² + 2n + 1

the square of n + d = n² + 2dn + d2

the square of n - 1 = n² - 2n +1

the square of n - d = n² -2dn + d2

the square of n - d = n² -2dn + d2

Where d is the difference between the number which you are trying to calculate the square of and, in the examples shown below, 5 and 10 respectively.

The idea is to pick some of the squares and remember them then the squares in between those numbers can be easily calculated. This way you will not have to remember the whole table above. Perhaps just as a suggestion you can remember the square of the fives and the tens and again some squares that you frequently encounter.
As example lets take square of 5 = 25 ( you should absolutely remember this one )
5² = 25
( add from 5 )

6² = 25 + 2 x 5 + 1 = 35

7² = 25 + 2x2x5 + 2x2 = 25+20+4=49

8² = 25 + 2x3x5 + 3x3 = 25 + 30 + 9 =64

9² = 25 + 2x4x5 + 4x4 = 25 + 40 + 16 = 81

or ( subtract from 10 )

8² = 10x10 - 2x10x2 + 2x2 = 100 - 40 + 4 = 64

9² = 10 x 10 - 2x10x1 + 1x1 = 100 - 20 + 1 = 81
I introduced this when I was dealing with the 6 Times table

This is efficient & fast and can be dealt with mentally.
Let’s review it a little more.

eg 6 x 8

6= 5 +1
8 = 5 + 3
Therefore, 6 x 8 = ( 5 + 1 ) x ( 5 + 3 )
the numbers are : 5 x 5 if you use this method always 25
and you really have only to calculate 2 numbers which are
( 3 + 1) x 5 = 4 x 5 = 20 and 1 x 3 = 3
6 x 8 = 25 + 20 + 3 = 48

eg. 8 x 3 = (5 + 3 ) x ( 5 - 2 )
= 25 + ( 3-2) x 5 + 3 x (-2)
= 25 + 5 - 6 = 24 ( This bothers on no deal because of the -ve number )

Hence only teach him/her to do it if both numbers are greater than 5.

6 x 7 = ( 5+ 1 ) x ( 5 + 2) = 25 + 15 + 2 = 42

Now let’s expand on this a little and apply it on the method I last introduced which is
multiplication using squares.

6 x 8
½ the sum of 6 & 8 = 14/2 = 7
6 + 8 = even & d = 8 - 7 = 1
hence 6 x 8 = v² - d² = 49 -1 = 48
I calculate the product by applying the procedure

6 x 7
v = 6
d = 1
sum = odd
6 x 7 = v² + v - d² + d = 6 x 6 + 6 - 1 x 1 + 1
= 36 + 6 + 0
= 42

7 x 4
v = 5
d = 1
odd
7 x 4 = 25 + 5 - 4 + 2 = 28

6 x 4
v = 5 , d = 1, even
6 x 4 = 25 - 1 = 24 ( v² - d²)