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5  Counting faster

5.1 Kangaroo math kangaroo

Important The Gigantic Turnip

“The Gigantic Turnip” is a cumulative or chain Russian folk tale. In a cumulative tale, the action repeats and builds in a cumulative manner.

The story is about an old man and an old woman who lived in a crooked cottage and owned the following animals:

  • 🐄
  • 🐖🐖
  • 🐈‍⬛🐈‍⬛🐈‍⬛
  • 🐓🐓🐓🐓
  • 🦢🦢🦢🦢🦢
  • 🐤🐤🐤🐤🐤🐤

Read more about this and other cumulative tales at https://wikipedia.org/

The numbers representing the number of each of these animals form an arithmetic progression, or arithmetic sequence, because the difference between any two consecutive numbers is constant. In this case, the common difference is 11.

In fact, the natural numbers N\color{red}{\mathbb{N}} are an arithmetic progression as well. As we saw before, the natural numbers can be generated by starting from 00 and repeatedly adding 11.

Our kangaroo kangaroo generates the natural numbers by jumping by ones.

There are two things you should know about our kangaroo kangaroo. First, his name is Joey, and he can jump more than one unit at a time. For example, he can jump two units at a time.

If we follow Joey, we get the sequence:

Joey can, of course, also jump in threes.

On the number line, we can skip the numbers that aren’t part of Joey’s jumps. The sequence remains the same; we are simply changing how we visualize it.

In both cases, the resulting sequence is the same:

Why is Kangaroo math important?

There are two important reasons to learn Kangaroo Math. First, it allows you to count faster. Second, it provides a solid foundation for learning multiplication, division, and exponentiation, as we will see in the next sections.

Counting by jumping two, three, or five units at a time allows you to count faster, just as climbing a staircase becomes quicker when you skip steps.

For example, you can climb a staircase twice as fast if you jump two steps at a time instead of one step at a time.

TipKangaroo math and multiplication

Multiplication is simply a shortcut for counting faster. We will learn later that there are even faster counting tricks.

Exercises

Let’s practice our Kangaroo Math! Write down the sequences generated when Joey jumps nn units at a time.

























Based on the following image, answer the following questions:

  • When you climb the stairs one step at a time, it takes you 1212 seconds to reach the top. If you jump 33 steps at a time at the same speed, how long would it take you to get to the top?
  • If you want to climb the stairs in 22 seconds, how many steps would you have to jump at a time?

5.2 Spotting patterns

As we saw earlier, multiplication is repeated addition—the special case that results from adding a value to itself multiple times. For example:

3+3+3+3+3⏟5 times=5×3=15\begin{aligned} \underbrace{3+3+3+3+3}_{{\color{red}{5}} \text{ times}} &= {\color{red}{5}} \times 3 = 15\\ \end{aligned}

Adding 33 to itself five times is the same as counting by threes five times.

That’s the connection between multiplication and Kangaroo Math!

To count faster, we must learn to do two related yet independent things.

  1. Spot a pattern in the objects we want to count.
  2. Practice and memorize—or rather, internalize—the results of common multiplications.

In the next section, we will start learning how to master multiplication results. For now, focus on spotting patterns in the objects you want to count.

Exercises

Try to identify the pattern in the following quantities, then write it down as a multiplication.

Here we have 55 rows of 33 bricks each. The total number of bricks is then 5×35 \times 3.

Here the total number of pink bricks is 5×3\boxed{\phantom{5}} \times \boxed{\phantom{3}}.

Here the total number of green bricks is 5×3\boxed{\phantom{5}} \times \boxed{\phantom{3}}.

There are 66 rows of 44 penguins each, plus an extra row with 33 penguins. Note that the last row cannot be included in the multiplication because it has a different number of penguins than the other rows.

There are 22 layers of 22 rows of 22 bricks each. The total number of bricks is then 2×2×22 \times 2 \times 2.

The total number of orange bricks is 5×3×2\boxed{\phantom{5}} \times \boxed{\phantom{3}} \times \boxed{\phantom{2}}.

The total number of yellow bricks is 5×3×2\boxed{\phantom{5}} \times \boxed{\phantom{3}} \times \boxed{\phantom{2}}.

Based on the chessboard below, answer the following questions. Can you express your answers using multiplication?

  • How many squares are there in total on the chessboard?
  • How many white pieces are on the chessboard?
  • How many pieces are there in total (white and black)?
  • How many squares are empty?

5.3 Multiplication tables

Important The Arabian Nights of Numbers

Beremiz Samir is a gifted mathematician who can count faster than anyone else. He can look at a flock of birds or a herd of camels and instantly tell you how many there are. His mathematical genius guides him through exciting and dangerous adventures as he solves puzzles and riddles, helps people, and finds love. Read more about Beremiz Samir in The Man Who Counted by Malba Tahan.

We can also learn to count quickly. As we saw in the previous section, the first step is identifying patterns. The second step is to practice and memorize the results of basic multiplication. Mastering these calculations will give you superpowers and lay the groundwork for algorithms for multiplying larger numbers.

We will use tables like the one below to practice our multiplication skills. Don’t worry about memorizing them; with practice, you will get better.

These tables are very similar to the ones we saw before for addition when defining groups.

The values in the table are the products of multiplying the row and column numbers. For example, to find the result of 3×43 \times 4, first locate the row with the number 33, then the column with the number 44. The cell where the row and column intersect contains the product, which is 1212 in this case. Therefore, 3×4=123 \times 4 = 12.

In the same way, we can compute for instance 2×52 \times 5.

If you look closely, you will see that the results of previous multiplications appear twice in the table.

Caution Why is that?

Stop and think!

Those values appear twice because multiplication, like addition, is commutative.

a×b=b×aa \times b = b \times a

In previous examples, we saw that 2×5=5×22 \times 5 = 5 \times 2 and 3×4=4×33 \times 4 = 4 \times 3. This property makes the multiplication table symmetric with respect to the diagonal.

What does it mean? It means good news! We only need to learn one half of the table: either the upper or lower triangle. Hooray! 🥳

However, keep in mind that the product you want to calculate may not be directly accessible, so you may need to use the commutative property to find it. For instance, if you want to calculate 5×25 \times 2, you could look for 2×52 \times 5 instead.

What are the sequences of each row and column?

By looking at one row or column in particular, such as the sequence of values in row nn, which is the same as the sequence of values in column nn, we obtain an arithmetic progression of multiples of nn. For example, the sequence of values in row 33 is the same as the sequence of values in column 33. We obtain the sequence 0,3,6,9,12,15,18,…0, 3, 6, 9, 12, 15, 18, \ldots.

In other words, the values in a given row or column correspond to the series in Kangaroo Math when Joey jumps nn units at a time.

What are the values in the diagonal?

Caution Question

What do the values in the diagonal of the multiplication table represent?

These are the results of multiplying a number by itself. We will soon learn that this operation is called squaring a number, and the result is called the square of the number.

a×a⏟2 times=a2\underbrace{a \times a}_{{\color{red}{2}} \text{ times}} = a^{{\color{red}{2}}}

For instance:

1×1=12=12×2=22=43×3=32=94×4=42=165×5=52=256×6=62=367×7=72=498×8=82=649×9=92=8110×10=102=100⋮\begin{align*} 1 \times 1 &= 1^{{\color{red}{2}}} = 1\\ 2 \times 2 &= 2^{{\color{red}{2}}} = 4\\ 3 \times 3 &= 3^{{\color{red}{2}}} = 9\\ 4 \times 4 &= 4^{{\color{red}{2}}} = 16\\ 5 \times 5 &= 5^{{\color{red}{2}}} = 25\\ 6 \times 6 &= 6^{{\color{red}{2}}} = 36\\ 7 \times 7 &= 7^{{\color{red}{2}}} = 49\\ 8 \times 8 &= 8^{{\color{red}{2}}} = 64\\ 9 \times 9 &= 9^{{\color{red}{2}}} = 81\\ 10 \times 10 &= 10^{{\color{red}{2}}} = 100\\ \vdots \\ \end{align*}

Exercises

Based on this multiplication table:

Calculate the following products:

2×7=1237×2=12311×11=12312×12=1235×9=1237×6=12311×1=127×1=1238×0=12312×0=123112=123122=123102=12342=123\begin{align*} 2 \times 7 &= \boxed{\phantom{123}}\\ 7 \times 2 &= \boxed{\phantom{123}}\\ 11 \times 11 &= \boxed{\phantom{123}}\\ 12 \times 12 &= \boxed{\phantom{123}}\\ 5 \times 9 &= \boxed{\phantom{123}}\\ 7 \times 6 &= \boxed{\phantom{123}}\\ 11 \times 1 &= \boxed{\phantom{12}}\\ 7 \times 1 &= \boxed{\phantom{123}}\\ 8 \times 0 &= \boxed{\phantom{123}}\\ 12 \times 0 &= \boxed{\phantom{123}}\\ 11^2 &= \boxed{\phantom{123}}\\ 12^2 &= \boxed{\phantom{123}}\\ 10^2 &= \boxed{\phantom{123}}\\ 4^2 &= \boxed{\phantom{123}}\\ \end{align*}

  • What happens when you multiply a number by 00?
  • What happens when you multiply a number by 11?
  • Write down the arithmetic series generated by counting in sevens.

The following are are the rules for multiplying negative numbers:

  • If the signs are the same (both positive or both negative), the result is positive. For example, (−2)×(−3)=6(-2) \times (-3) = 6 and 2×3=62 \times 3 = 6.
  • If the signs are different (one positive and one negative), the result is negative. For example, (−2)×3=−6(-2) \times 3 = -6 and 2×(−3)=−62 \times (-3) = -6.

In other words, the sign is the exclusive or (XOR), or the modulo 22 sum of the signs of the factors.

Caution Question

Why multiplying two negative numbers results in a positive number?

2×−7=−147×−2=−14−11×−11=12312×12=1235×−9=−457×−6=12311×−1=12−7×−1=1232×0=123−2×0=123(−10)2=123(−4)2=123\begin{align*} 2 \times -7 &= \boxed{\phantom{-14}}\\ 7 \times -2 &= \boxed{\phantom{-14}}\\ -11 \times -11 &= \boxed{\phantom{123}}\\ 12 \times 12 &= \boxed{\phantom{123}}\\ 5 \times -9 &= \boxed{\phantom{-45}}\\ 7 \times -6 &= \boxed{\phantom{123}}\\ 11 \times -1 &= \boxed{\phantom{12}}\\ -7 \times -1 &= \boxed{\phantom{123}}\\ 2 \times 0 &= \boxed{\phantom{123}}\\ -2 \times 0 &= \boxed{\phantom{123}}\\ (-10)^2 &= \boxed{\phantom{123}}\\ (-4)^2 &= \boxed{\phantom{123}}\\ \end{align*}

5.4 Exponentiation

As we saw earlier, repeated addition is multiplication. For example, 4×74\times 7 is equivalent to adding 77 four times: 7+7+7+77+7+7+7. In the general case, we have:

a+a+⋯+a⏟n times=n×a\begin{aligned} \underbrace{a + a + \cdots + a}_{{\color{red}{n}} \text{ times}} &= {\color{red}{n}} \times a \end{aligned}

Because mathematics is recursive, much like the Gigantic Turnip, we can now ask ourselves the following question:

Caution Question

If repeated addition is multiplication, then what is repeated multiplication?

As we’ve seen, multiplying a number by itself is called squaring the number. We symbolize this with a cute little superscript.

a×a=a2a \times a = a^2

This little number represents how many times we multiply a number by itself. It is called the exponent.

a×a×⋯×a⏟n times=an\begin{aligned} \underbrace{a \times a \times \cdots \times a}_{{\color{red}{n}} \text{ times}} &= a^{\color{red}{n}} \end{aligned}

Just as we call it “squaring” when the exponent is two, we call it “cubing” when the exponent is three because we are defining the volume of a cube.

Here are some examples of exponentiation:

22=2×2=423=2×2×2=842=4×4=1633=3×3×3=27\begin{align*} 2^2 &= 2 \times 2 &= 4 \\ 2^3 &= 2 \times 2 \times 2 &= 8 \\ 4^2 &= 4 \times 4 &= 16 \\ 3^3 &= 3 \times 3 \times 3 &= 27 \end{align*}

5.5 Exponentiation of 10

A special case of exponentiation occurs in the decimal system when the base is 1010.

101=10=10102=10×10=100103=10×10×10=1000104=10×10×10×10=10000\begin{align*} 10^1 &= 10 &= 10 \\ 10^2 &= 10 \times 10 &= 100 \\ 10^3 &= 10 \times 10 \times 10 &= 1000 \\ 10^4 &= 10 \times 10 \times 10 \times 10 &= 10000 \end{align*}

In general we have:

10×10×⋯×10⏟n times=10n1000⋯000⏟n zeros\begin{aligned} \underbrace{10 \times 10 \times \cdots \times 10}_{{\color{red}{n}} \text{ times}} &= 10^{\color{red}{n}} & 1\underbrace{\color{red}{000\cdots 000}}_{{\color{red}{n}} \text{ zeros}} \end{aligned}

Below is a list of the first powers of 1010:

100=1101=10102=100103=1000104=10000105=100000\begin{align*} 10^0 &= 1 \\ 10^1 &= 10 \\ 10^2 &= 100 \\ 10^3 &= 1000 \\ 10^4 &= 10000\\ 10^5 &= 100000 \\ \end{align*}

Notice that 10010^0 is equal to 11. This is true for any number, not just 1010. a0=1a^0=1.

As I mentioned before, mathematics is recursive.

Caution Question

In the same way that repeated addition is multiplication and repeated multiplication is exponentiation, what is repeated exponentiation?

We will stop at exponentiation. As an interesting fact, repeated exponentiation is called tetration, which allows us to build towers of numbers like this:

23332^{3^{3^3}}

Exercises

Calculate the following powers:

23=824=1632=9100=1101=10105=100000104=10000\begin{align*} 2^3 &= \boxed{\phantom{8}} \\ 2^4 &= \boxed{\phantom{16}} \\ 3^2 &= \boxed{\phantom{9}} \\ 10^0 &= \boxed{\phantom{1}} \\ 10^1 &= \boxed{\phantom{10}} \\ 10^5 &= \boxed{\phantom{100000}} \\ 10^4 &= \boxed{\phantom{10000}} \\ \end{align*}

5.6 Positional number system revisited

Just as calculating powers of 1010 is easy, so too is multiplying a digit by a power of 1010.

3×102=3×100=3007×103=7×1000=70002×104=2×10000=200003×105=3×100000=300000\begin{align*} 3 \times 10^2 &= 3 \times 100 = 300 \\ 7 \times 10^3 &= 7 \times 1000 = 7000\\ 2 \times 10^4 &= 2 \times 10000 = 20000 \\ 3 \times 10^5 &= 3 \times 100000 = 300000 \end{align*}

Do you see the pattern?

Now that we have this information, we can finally formalize the positional number system.

Remember that in a positional number system, the value of a digit depends on its position.

Caution Question

But what is the value of each digit?

3⏟23⏟13⏟0\underbrace{\color{blue}{3}}_{2}\underbrace{\color{red}{3}}_{1}\underbrace{\color{Black}{3}}_{0}

The same number can be written as:

333=3×102+3×101+3×100{\color{Blue}{3}}{\color{Red}{3}}{\color{Black}{3}} = {\color{Blue}{3}} \times 10^2 + {\color{Red}{3}} \times 10^1 + {\color{Black}{3}} \times 10^0

Therefore, we can write the same number as the sum below:

300+30+3333\begin{aligned} {\color{Blue}{3}}00 \\ +{\color{Red}{3}}0 \\ +{\color{Black}{3}}\\ \hline {\color{Blue}{3}}{\color{Red}{3}}{\color{Black}{3}} \end{aligned}

Here is another example:

150=1×102+5×101+0×100{\color{Blue}{1}}{\color{Red}{5}}{\color{Black}{0}} = {\color{Blue}{1}} \times 10^2 + {\color{Red}{5}} \times 10^1 + {\color{Black}{0}} \times 10^0

4000+100+00+14101\begin{aligned} {\color{OliveGreen}{4}}000 \\ +{\color{Blue}{1}}00 \\ +{\color{Red}{0}}0 \\ +{\color{Black}{1}}\\ \hline {\color{OliveGreen}{4}}{\color{Blue}{1}}{\color{Red}{0}}{\color{Black}{1}} \end{aligned}

In general, the value of each digit did_i is given by:

di×10pid_i \times 10^{p_i}

Where pip_i is the position of the digit did_i in the number, starting at 00 for the rightmost digit.

TipFun Fact

This holds true for any base, not just 1010, and can be extended to non-integer numbers as well.

Exercises

Write each of the following numbers as its constituent parts:

123=100+20+3100=100+0+04567=4000+500+60+717=10+123=20+31001=1000+0+0+11000=1000+0+0+0555=500+50+5171=100+70+179=70+942=40+210381=10000+0+300+80+1\begin{align*} 123 &= 100 + 20 + 3 \\ 100 & = 100 + 0 + 0 \\ 4567 &= 4000 + 500 + 60 + 7 \\ 17 &= \boxed{\phantom{10}} + \boxed{\phantom{1}} \\ 23 &= \boxed{\phantom{20}} + \boxed{\phantom{3}} \\ 1001 &= \boxed{\phantom{1000}} + \boxed{\phantom{0}} + \boxed{\phantom{0}} + \boxed{\phantom{1}}\\ 1000 &= \boxed{\phantom{1000}} + \boxed{\phantom{0}} + \boxed{\phantom{0}} + \boxed{\phantom{0}}\\ 555 &= \boxed{\phantom{500}} + \boxed{\phantom{50}} + \boxed{\phantom{5}}\\ 171 &= \boxed{\phantom{100}} + \boxed{\phantom{70}} + \boxed{\phantom{1}}\\ 79 &= \boxed{\phantom{70}} + \boxed{\phantom{9}}\\ 42 &= \boxed{\phantom{40}} + \boxed{\phantom{2}}\\ 10381 &= \boxed{\phantom{10000}} + \boxed{\phantom{0}} + \boxed{\phantom{300}} + \boxed{\phantom{80}} + \boxed{\phantom{1}}\\ \end{align*}

We saw in section Section 1.8 how to represent values using a single-column Soroban abacus.

Here are some examples to refresh your mamory:

The Soroban abacus follows the same logic as the positional number system, meaning that the first column to the right represents units, that is the value multiplied by 10010^0, the next column represents tens, that is the value multiplied by 10110^1, the next column represents hundreds, that is the value multiplied by 10210^2, and so on.

Complete the following representations:

5.7 Division

Imagine that you are playing a card game with your friends. It could be Uno, The Mind, or any other card game in which each player starts with an equal number of cards.

Let’s say the game has 1212 cards and 33 are three players. Initially, each player has 00 cards.

Start by giving each player one card at a time. After the first round, each player will have 11 card.

After the second round, each player will have 22 cards.

After the third round, each player will have 33 cards.

This continues until all cards are distributed.

By the end, every player has 44 cards.

This process of splitting into equal parts is called division. In the previous example, we said that 1212 cards divided among 33 players equals 44 cards per player. There are different ways to write this.

12÷3=12/3=123=412 \div 3 = 12/3 = \frac{12}{3} = 4

Caution Question

How many cards will each player get if you invite a new friend and play the same game with 44 players?

If we follow the same process of giving out the cards one by one, each player ends up with 33 cards.

This can be written as follows:

12÷4=12/4=124=312 \div 4 = 12/4 = \frac{12}{4} = 3

What is division?

Just as subtraction is the inverse of addition, division is the inverse of multiplication.

To see this, let’s review the previous results and try to connect them with multiplication. We can write the first result as follows:

123=412=3×4\begin{align*} \frac{12}{3} &= 4 \\ 12 &= 3 \times 4 \end{align*}

For example, we determined that there are 5×3=155\times 3 = 15 LEGO bricks in the following image:

We can also work backwards and ask, given that there are 1515 LEGO bricks, how many rows of 33 bricks can we build? 15/3=515/3 = 5. How many columns of 55 bricks can we build? 15/5=315/5 = 3.

In general, if an integer cc is the product of two integers aa and bb,

c=a×bc = a \times b

We known than cc can be divided by both aa and bb:

c÷b=ac÷a=b\begin{align*} c \div b &= a \\ c \div a &= b \end{align*}

This means that we can use multiplication tables to find division results.

For example, if I asked you to solve the following division problem:

355=?\frac{35}{5} = \boxed{?}

We can look at the multiplication table to find the values where the product is 3535. Then, we can check if any of the columns or rows of those values contains a 55.

Since 3535 is the product of 55 and 77, we can conclude that 35÷5=735 \div 5 = 7.

Exercises

  • If I have 2020 LEGO bricks, how many towers of 44 bricks each can I build?
  • If my grandparents bring me and my sister 88 bombons, and we want to split them equally, how many will each of us get?
  • Compute the following divisions using the multiplication table:

36÷6=648÷8=656÷7=863÷9=772÷8=9100÷10=10\begin{align*} 36 \div 6 &= \boxed{\phantom{6}} \\ 48 \div 8 &= \boxed{\phantom{6}} \\ 56 \div 7 &= \boxed{\phantom{8}} \\ 63 \div 9 &= \boxed{\phantom{7}} \\ 72 \div 8 &= \boxed{\phantom{9}} \\ 100 \div 10 &= \boxed{\phantom{10}} \\ \end{align*}

  • In the following games, how many cards does each player get?
21 cards divided by 3 players:

20 cards divided by 4 players:

15 cards divided by 5 players:

9 cards divided by 3 players:

5.8 Multiplying and dividing by powers of 10

Multiplying and dividing by powers of 1010 is a special case that can be done quickly by shifting the digits of a number. This is because 1010 is the base of our decimal number system. The same idea works in any base and is how computers efficiently perform multiplication and division by powers of 22.

Given a number with multiple digits, d1d2…dn,d_1d_2\dots d_n, and a power of 1010 to multiply by, the result is simply the original number shifted to the left by the exponent of 1010 (the number of zeros in the power of ten) for multiplication and to the right for division.

Remember, a power of 1010 is a 11 followed by 0s0\text{s}.

10×10×⋯×10⏟n times=10n1000⋯000⏟n zeros\begin{aligned} \underbrace{10 \times 10 \times \cdots \times 10}_{{\color{red}{n}} \text{ times}} &= 10^{\color{red}{n}} & 1\underbrace{\color{red}{000\cdots 000}}_{{\color{red}{n}} \text{ zeros}} \end{aligned}

This is what happens when we multiply a general number d1d2…dnd_1d_2\dots d_n by a power of 1010:

d1d2…dn⏟number×10n=d1d2…dn⏟number×10⋯0⏟n zeros=d1d2…dn0⋯0⏟number with n zeros at the end\begin{aligned} \underbrace{d_1d_2\dots d_n}_{\text{number}} \times 10^{\color{red}{n}} &= \underbrace{d_1d_2\dots d_n}_{\text{number}} \times \underbrace{1\color{red}{0\cdots 0}}_{{\color{red}{n}} \text{ zeros}} \\ &= \underbrace{d_1d_2\dots d_n\color{red}{0\cdots 0}}_{\text{number with $n$ zeros at the end}} \end{aligned}

The following examples should make this concept clearer.

5×101=5×10=505×102=5×100=500753×104=753×10000=7530000\begin{aligned} {\color{OliveGreen}5} \times 10^{\color{red}1} &= {\color{OliveGreen}5} \times 1{\color{red}0} = {\color{OliveGreen}5}{\color{red}0} \\ {\color{OliveGreen}5} \times 10^{\color{red}2} &= {\color{OliveGreen}5} \times 1{\color{red}00} = {\color{OliveGreen}5}{\color{red}00} \\ {\color{OliveGreen}753} \times 10^{\color{red}4} &= {\color{OliveGreen}753} \times 1{\color{red}0000} = {\color{OliveGreen}753}{\color{red}0000} \end{aligned}

Exercises

Perform the following computations:

2×10=2020×10=200123×1000=123000123×103=12300057×102=570021×100=21001000÷10=1001200÷100=125700÷10=5705700÷102=57\begin{align*} 2 \times 10 &= \boxed{\phantom{20}}\\ 20 \times 10 &= \boxed{\phantom{200}}\\ 123 \times 1000 &= \boxed{\phantom{123000}}\\ 123 \times 10^3 &= \boxed{\phantom{123000}}\\ 57 \times 10^2 &= \boxed{\phantom{5700}}\\ 21 \times 100 &= \boxed{\phantom{2100}}\\ 1000 \div 10 &= \boxed{\phantom{100}}\\ 1200 \div 100 &= \boxed{\phantom{12}}\\ 5700 \div 10 &= \boxed{\phantom{570}}\\ 5700 \div 10^2 &= \boxed{\phantom{57}}\\ \end{align*}