We’ll start with the easy stuff.

(1) Double the multiplicand you want to multiply 4 by

(2) Double it one more time

e.g. 8 * 4 = 32

**Multiplying by 4:**(1) Double the multiplicand you want to multiply 4 by

(2) Double it one more time

e.g. 8 * 4 = 32

8 * 2 = 16

16 * 2 = 32

*Why does this work?*

4 can be broken up into 2 * 2

8 * 4 = 32

8 * (2 * 2) = 32

Thanks to the associative property of multiplication, we can multiply factors in whatever grouping or order we chose and still get the same answer. We start by multiplying multiplicand we want to multiply 4 by 2 because this computation is easy for most people to do in their heads.

(8 * 2) * 2

(16) * 2

We then multiply our product by the remaining multiplicand, which is 2.

16 * 2 = 32

**Multiplying by 10:**

Stick a zero behind whatever number you wish to multiply 10 by

988 * 10

= 9880

*Why does this work?*

Consider what we’re doing in terms of place value. When multiplying a number by 10, we’re essentially moving everything up a place value.

10 ones = 10

10 tens = 100

10 hundreds = 1,000

10 thousands = 10,000

etc.

Going back to our original example of 988 x 10 we see that:

8 ones x 10 becomes 8 tens or 80

8 tens x 10 becomes 8 hundreds or 800

9 hundreds x 10 becomes 9 thousands or 9,000

Let’s take a look at the process again, this time using long multiplication:

988

x 10

80 We’re multiplying each digit first by zero ones, then by one ten.

800 That combination keeps getting pushed over as you multiply by larger and larger

+9000 places so the original digits are simply moved over one place.

9880

You probably knew those already. Let’s move on to some lesser known ones:

(1) Multiply the factor you want to multiply 5 by by 10

(2) Divide that product by 2

66 * 5

66 * 10 = 660

660/2 = 330

That 10 trick was pretty easy, right? Well, the five trick is basically a modified version of that.

We multiply by 10 because it’s pretty easy to do in our heads. Then we divide by 2 because 5 is half of 10 so our final product must be half of our product from step one.

9880

You probably knew those already. Let’s move on to some lesser known ones:

**Multiplying by 5:**(1) Multiply the factor you want to multiply 5 by by 10

(2) Divide that product by 2

66 * 5

66 * 10 = 660

660/2 = 330

*Why does this work?*That 10 trick was pretty easy, right? Well, the five trick is basically a modified version of that.

We multiply by 10 because it’s pretty easy to do in our heads. Then we divide by 2 because 5 is half of 10 so our final product must be half of our product from step one.

**Multiplying by 9 (only works for 9x1 - 9x10):**

(1) Spread your fingers out in front of you (doesn’t matter whether palms are facing up or down)

(2) Going from left to right, fold the finger down that you wish to multiply 9 by (if you wish to multiply 9 * 3 then fold down your middle finger on your left hand)

(3) Everything to the left of your folded down finger represents the tens digit and everything to the right represents the ones digit

*Why does this work?*

Notice that the sum of all the digits of numbers divisible by 9 is equal to or divisible by 9

3 * 9 = 27

2 + 7 = 9

8 * 9 = 72

7 + 2 = 9

This is true of all numbers divisible by 9.

5418/9 = 602

5 + 4 = 9

1 + 8 = 9

999963/9 = 111107

9 = 9

9 = 9

9 = 9

9 = 9

6 + 3 = 9

But how does this work? We need to rethink how we think about numbers:

(1) Take the number that you’re interested in seeing if it’s divisible by 9 and split it up into its places.

5418 becomes

5(1000) + 4(100) + 1(10) + 8(1)

(2) If n represents a place, rewrite what’s in the parentheses to ((n - 1) + 1)

5(999 + 1) + 4(99 + 1) + 1(9 + 1) + 8

How are we allowed to do this? Well, you’ll see that if you add everything up and then multiply, you’ll still get 5418. We’re simply expressing the number differently.

(3) Distribute

5 + (5 * 999) + 4 + (4 * 99) + 1 + (1 * 9) + 8

(4) We know that everything within the parenthesis is divisible by 9 because we are multiplying what’s inside by a multiple of nine. So that just leaves that string on the end, which added together also has to be divisible by 9. Notice anything interesting about them? Yep, they are our original digits!

(5 * 999) + (4 * 99) + (1 * 9) + 8 + 5 + 4 + 1

8 + 5 + 4 + 1 = 18

1 + 8 = 9

**Multiplying by 11:**

(1) Erase all numbers in between the ones digit and the largest place value of the multiplicand you wish to multiply 11 by (e.g. 54321 becomes 5 _ _ _ 1). Skip to step two if you’re multiplying a two digit number.

(2) Add an additional space in between the ones and the largest place

(5 _ _ _ _1)

(3) Add together the ones digit and the tens digit from the multiplicand that you wish to multiply 11 by and insert the sum into the right-most blank place.

543

(4) Working from right to left, move one place over so now you’re working with the tens and hundreds digits. Add those two together and insert the sum into the right-most empty spot.

54

(5) Keep moving through the places until you run out of numbers.

5

And there you have it!

What happens when the sum in step two is greater than 9 - i.e. how to carry the 1?

Simple! You carry the one over to the next place.

157 * 11

1_7

1_ _ 7

1 _ 27 because 7 + 5 = 12

1727 you’d think it’d be 6 because 1 + 5 = 6 HOWEVER, even though we usually don’t mess with the largest place, we still have that one leftover from the previous step. Thus, the hundreds place becomes 7 and the 1 gets bumped to the thousands place in this case.

This “trick” is a lot more straightforward than you’d think. When you do the long multiplication, you see that you’re simply adding up adjacent digits

54321

x 11

11

220

3300

44000

543

**21**2 + 1 = 3 5 _ _ _ 31(4) Working from right to left, move one place over so now you’re working with the tens and hundreds digits. Add those two together and insert the sum into the right-most empty spot.

54

**32**1 3 + 2 = 5 5 _ _ 531(5) Keep moving through the places until you run out of numbers.

5

**43**21 4 + 3 = 7 5 _ 7531**54**321 5 + 4 = 9 597531And there you have it!

What happens when the sum in step two is greater than 9 - i.e. how to carry the 1?

Simple! You carry the one over to the next place.

157 * 11

1_7

1_ _ 7

1 _ 27 because 7 + 5 = 12

1727 you’d think it’d be 6 because 1 + 5 = 6 HOWEVER, even though we usually don’t mess with the largest place, we still have that one leftover from the previous step. Thus, the hundreds place becomes 7 and the 1 gets bumped to the thousands place in this case.

*How does this work?*This “trick” is a lot more straightforward than you’d think. When you do the long multiplication, you see that you’re simply adding up adjacent digits

54321

x 11

11

220

3300

44000

+550000

597531

597531