Arithmetic with Playing Cards

Best Training Aid

In my opinion,the playing cards, is the best system to train young- schoolers in Numbers and Arithmetic for all of the followings:

• It is cheap and easily acquirable
• It can be used almost anywhere with minimum space
• It is fun to use
• Its usage starts when kids are just learning to count
• It is a tool for training on Heuristics

Preliminaries:

Usage of the cards is simplistic. Treat the picture cards as tens or a set as Universal Cards.
Universal cards are used as cards of a particular value if shortage of cards of the value occurs.
Jokers are used as Universal cards. You get to set the rules for your charge.
Although I am recommending the playing cards I use different visual cards with pictures of animals
or some other famous characters out of movies for a couple of reasons:
1) Provide a way to keep attention focused
2) Allow people who , for understandable reasons, hold beliefs which are not in favor of using playing cards

Addition

7 + 8 = ?
When I use the playing cards I will draw out a 7 and a 8 .

Have your charge count through each of the three cards one at a time
After each count compare finger work against the pictures.
Hopefully your charge will end up with 2 hands and asking for help.
This is when you ask the kid to close 1 hand and use it to count
from 11 to 15 .

This is the only example I will put up because this document is not meant
for this - my assumptions were that the Junior schoolers would have acquired
the concept of numbers and the ability to count through them. If there is
really a need for such it can be considered for a subsequent blog, but I doubt
that, as the resources for such topic have flooded the Internet and in the
Android as well as Microsoft and IOS stores. Hence on I will use the cards
to illustrate the 4 Arithmetic Operations.

Adding 7 + 8

Whenever performing Addition and Multiplication we try to create as many values of 10 as we can.
Each ten created brings you closer to the correct answer. Sometimes it takes only 1 or 2 tens
to know the answer of the operation.

= ?

We split the 8 into 5 and 3. The 3 then go on to form a 10 with 7.

=

15

An Alternative way to perform this addition.

The seven is split into 5 and 2.
The 2 is then combined with the 8 to form a 10

A third way of performing this addition.

Add 3 to 7 and 2 to 8 in essence borrowing 2 and 3.
The borrowed numbers are then subtracted from the intermediate sum.
In this case we have 10 + 10 = 20
And 20 - 5 = 15 (the final answer)

Noticed that between the 2 cards there are 5 icons painted red.
This is to identify the values as borrowed. The values must be subtracted
after the fact - in this case 20.
Hence Answer = 20 - 5 = 15

This third method is very conducive to mental Arithmetic

Another Example - 37 + 25

The 5 10(s) do not require additional mention so we will deal with the 7 and the 5.

+

=

+

+

=

12

37+25=50+12

=62

Subtraction

Before we delve into subtraction there are a couple of things to note:
1) The value of a number is the sum of the values of all the cards
2) Each card has a value as stored on its face and it does not have a positional
value. Meaning that the value of the card does not change regardless of where
it is positioned.

The value of sum of both cards is 11 and not 38.
Even if the 2 cards are switched in position their combined value would still be 11.
If there is any weakness in this system it got to be this. In the real world we use
the decimal system and the value of a number is dependent on its position.

Subtracting 4 from 9

Show a 9 and cover four icons

= 5

Example2 : 37 - 19

The 2 cards that form 19 are colored red to denote the amount to subtract.
The 10 from 19 is remove by using one of the tens from 37.

The remaining 9 is then subtracted from one of the tens from
the remaining 27.

-

=

Since there is no more number to subtract
what is left to do is to sum up the remaining numbers which
are 10 + 7 + 1 = 18

Answer : 18

Subtraction Techniques 37-19

Multiplication

Multiplication is repeated Addition.
7 x 4 = 4 + 4 + 4 + 4 + 4 + 4 + 4
4 x 7 = 7 + 7 + 7 + 7
Both = 28

7 x 4 = ?

Every 3 x 4 = 12. Therefore we replace every 3 x 4 cards with 1 x 10 and 1 x 2 cards

=

= 28

7 x 8 = ?

This is an interesting example.
There are 4 ways to solve this multiplication.

Method 1

7 x 8 = 8 + 8 + 8 + 8 +8 + 8 + 8

Every 8 requires a 2 to make 10. Therefore if we break up an 8 into 2(s)
we will have 4 10(s) and left with 2 8(s).
If we then break up 1 of the 8(s) into 6 and 2
we get another 10 with 6 remaining.

Answer: 7 x 8 = 56

Method 2

7 x 8 = 10 x 8 - 3 x 8
It is likely that your charge may not have learned 3 x 8.
However, I strongly believe that he or she would have been
taught that 3 x 8 = 8 x 3.

The 3 8(s) in red are there to remind us that their values must be deducted

When I use the playing cards, I would lay down 10 8(s) if I have that many and if not I would substitute with Jokers - telling my charge that the Jokers are used as 8(s). I would use red Jacks to represent the 3 red 8(s)

The 10 8(s) make 80 and the 3 8(reds) make 24

I would use 3 of the 10(s) to subtract the 24 leaving a 6 which should be returned to the rest of the cards.

Method 3

7 x 8 = 10 x 8 - 3 x 8

3 x 8 = 8 x 3 = 24 and 10 x 8 = 80
Hence we end up with 80 - 24

80 - 24 = 56

Method 4

7 x 8 = 5 x 8 + 2 x 8

5 x 8 = 8 x 5
8 is even and half of 8 = 4
hence, 8 x 5 = 40 and 2 x 8 = 16
40 + 16 = 56

Multiplication By 5

Note: There are some tables viz, 1,10,11 and 5 that do not require
rote learning (in my opinion.) Therefore I won't even bother to use
the card system on them.

Division

Division is repeated Subtraction.
Although a solid foundation on Multiplication will render Division a walk in the breeze, however
Division can be perform without the use of Multiplication.

When we perform the division of a number ( known as the dividend ) by another number ( known as the divisor ) we are trying to find out how many groups of the divisor can be derived from the dividend. I know I have thrown big terms but fear not. These are not terms that will be examined but suffice that when others mention the terms you know what they mean.

To make things clear let consider 6 ÷ 2
6 ÷ 2 = how many groups of 2 can be created from 6 ?
Since 6 = 2 + 2 + 2 the answer is 3.
In this example, 6 is the dividend, 2 is the divisor and the quotient is 3
The next question is, "How is Division related to Subtraction ?"

6 ÷ 2

​

6	Amount deducted	Number of group
6	0	0
4	2	1
2	2	1
0	2	1

The table above illustrate repeated Substraction
As can be concluded from the table, if the number to be divided is big then repeated subtraction can take "forever".

Multiplication vs Division

Two numbers when multiplied together forms a new number ( known as the product ).
If, number1 x number2 = product ( Multiplication)
then product ÷ number1 = number2
and,
product ÷ number2 = number1
Hence, we can and we will apply this understanding to determine the result of a Division.
Example 12 ÷ 3 = ?
Since 12 = 3 x 4
therefore 12 ÷ 3 = 4

Next Example : 48 ÷ 6 = ?

Lets start by laying down 4 x 10 + 1 x 8 on the screen

We are dividing 48 by 6, therefore we have to determine the number of groups of 6 that can be formed from 48. I start by noting that 10 = 6 + 4
and 8 = 6 + 2

The next logical step is to convert the 10(s) and 8 into 6 and we end up with this.

Combining the 2 with a 4 gives a 6. One of the 3 remaining 4 is broken up into 2 2(s) and which are then combined with the remaining 2 4(s) to create 2 6(s)

The result of the manipulation of the cards result in 8 6(s). Each of the card (6) represents a group of 6. Hence the result of dividing 48 by 6 is 8

48 div 6 via Multiplication

48 = 8 x 6 = 6 x 8
Then, 48 div 6 = 8 and 48 div 8 = 6
Since we are performing 48 div 6 then it means that 8 is the solution

Using the card system to perform Division
by making use of repeated-subtraction

This time we perform 84 div 7
Since its 84 div 7 we just perform 84 - 7 = 77 - 7 ....... until 84 is wilted down to 0 or less than 7.
The way we handle the cards is this: Start by laying 10 8(s) and 1 four

We then remove a 10 but replace it with a 7, into another group ( quotient-group ), but return a 3.

4 + 3 = 7. Therefore we remove both cards and add a seven to the quotient pile. If it is not obvious, we in in effect subtracting 7 from 84 at each step. Each time we place a 7 in the quotient pile to show that a group of 7 has been formed.

In the next subsequent 3 steps we are just subtracting 7 from 10 each time adding a 7-card to the quotient stack and returning a 3 to the dividend.

At this stage the answer is already hidden in plain site. However for the sake of completing this algorithm I will soldier on.

Hint : both stacks add up to 42 , therefore the solution is just twice the number of 7(s) on the quotient stack

Next we subtract 7 from 10 twice, add 2 7(s) to the quotient stack and return 2 3(s) to the dividend.

This step is rather obvious. We subtract 7 from 8 ( 3 + 3 + 2 ), put 1 more 7-card on the quotient stack and return 1 to the dividend.

After we subtract the 2 10(s) ,return 2 3(s) to the dividend and put 2 7(s) onto the quotient stack we end up with this picture.

This is really the last step since ( 3 + 3 + 1 = 7 ), put 1 more 7-card on the quotient stack and we are done.

The End

Since there are 12 groups of 7 the answer to 84 div 7 = 12

Though the Card system is very powerful, nevertheless this example shows the inefficiency of Repeated-Subtraction. This is why the genius of old invented Division, particularly Long Division. Nevertheless if you put thoughts on the matter ( if not read up my blog on Division ) this system can overcome such inefficiency.
Actually, this example is a trivial one. If your charge had paid attention to my exhortation on the 11-Time-Table he/she would have known that 11 x 7 = 77 and hence 12 x 7 = 84 and with that
would have solve this exercise in mere seconds.