Power Play Class 8 Notes Maths Chapter 2

Students often refer to Class 8 Maths Notes and Chapter 2 Power Play Class 8 Notes during last-minute revisions.

Class 8 Maths Chapter 2 Notes Power Play

Class 8 Maths Notes Chapter 2 – Class 8 Power Play Notes

→ We analysed some situations, asked questions, and found answers by first guessing, then modelling the problem statement, followed by making assumptions and approximations to carry out the calculations.

→ We experienced how rapid exponential growth, also called multiplicative growth, can be compared to additive growth.

→ n^a is n × n × n × n × ..… × n (n multiplied by itself a times) and \(n^{-a}=\frac{1}{n^a}\).

→ Operations with exponents satisfy

• n^a × n^b = n^a+b
• (n^a)^b = (n^b)^a = n^a×b
• n^a ÷ n^b = n^a-b (n ≠ 0)
• n^a × m^a = (n × m)^a
• n^a ÷ m^a = (n ÷ m)^a (m ≠ 0)
• n⁰ = 1 (n ≠ 0)

→ The scientific notation for the number 308100000 is 3.081 × 10⁸.

→ The standard form of the scientific notation of any number is x × 10^y, where x ≥ 1 and x < 10, and y is an integer.

→ Engaging in interesting thought experiments can be used as a means to understand how large a number or a quantity is.

Experiencing the Power Play… Class 8 Notes

An Impossible Venture!
Take a sheet of paper, as large a sheet as you can find. Fold it once. Fold it again, and again.

How many times can you fold it over and over?
Estu says, “I heard that a sheet of paper can’t be folded more than 7 times”.
Roxie replies, “What if we use a thinner paper, like a newspaper or a tissue paper?”
Try it with different types of paper and see what happens.

Say you can fold a sheet of paper as many times as you wish. What would its thickness be after 30 folds? Make a guess.
Let us find out how thick a sheet of paper will be after 46 folds. Assume that the thickness of the sheet is 0.001 cm.

The following table lists the thickness after each fold. Observe that the thickness doubles after each fold.

(We use the sign ‘≈’ to indicate ‘approximately equal to’.)
After 10 folds, the thickness is just above 1 cm (1.024 cm).
After 17 folds, the thickness is about 131 cm (a little more than 4 feet).
Now, what do you think the thickness would be after 30 folds? 45 folds? Make a guess.

After 26 folds, the thickness is approximately 670 m. Burj Khalifa in Dubai, the tallest building in the world, is 830 m tall.

After 30 folds, the thickness of the paper is about 10.7 km, the typical height at which planes fly. The deepest point discovered in the oceans is the Mariana Trench, with a depth of 11 km.

It might be hard to digest the fact that after just 46 folds, the thickness is more than 7,00,000 km. This is the power of multiplicative growth, also called exponential growth. Let us analyse the growth. We have seen that the thickness doubles after every fold.

Notice the change in thickness after two folds. By how much does it increase? After any 3 folds, the thickness increases 8 times (= 2 × 2 × 2). Check if that is true. Similarly, from any point, the thickness after 10 folds increases by 1024 times (= 2 multiplied by itself 10 times), as shown in the table below.

Exponential Notation and Operations Class 8 Notes

The initial thickness of the paper was 0.001 cm.
Upon folding once, its thickness became 0.001 cm × 2 = 0.002 cm.
Folding it twice, its thickness became 0.001 cm × 2 × 2 = 0.004 cm, or 0.001 cm × 2² = 0.004 cm (in shorthand).
Upon folding it thrice, its thickness became 0.001 cm × 2 × 2 × 2, or 0.001 cm × 2² = 0.008 cm.
When folded four times, its thickness became 0.001 cm × 2 × 2 × 2 × 2, or 0.001 cm × 2⁴ = 0.016 cm.
Similarly, the expression for the thickness of the paper when folded 7 times will be 0.001 cm × 2 × 2 × 2 × 2 × 2 × 2 × 2, or 0.001 cm × 2⁷ = 0.128 cm.
We have seen that square numbers can be expressed as n² and cube numbers as n³.
n × n = n² (read as ‘n squared’ or ‘n raised to the power 2’)
n × n × n = n³ (read as ‘n cubed’ or ‘n raised to the power 3’)
n × n × n × n = n⁴ (read as ‘n raised to the power 4’ or ‘the 4th power of n’)
n × n × n × n × n × n × n = n⁷ (read as ‘n raised to the power 7’ or ‘the 7th power of n’) and so on.
In general, we write na to denote n multiplied by itself a times.
5⁴ = 5 × 5 × 5 × 5 = 625.
5⁴ is the exponential form of 625.
Here, 4 is the exponent/power, and 5 is the base.
Exponents of the form 5ⁿ are called powers of 5: 5¹, 5², 5³, 5⁴, etc.
2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2¹⁰ = 1024.
Remember the 1024 from earlier? There, it meant that after every 10 folds, the thickness increased 1024 times. 54 is read as ‘5 raised to the power 4’ or ‘5 to the power 4’ or ‘5 power 4’ or ‘4th power of 5’

Some more examples of exponential notation:
4 × 4 × 4 = 4³ = 64.
(-4) × (-4) × (-4) = (-4)³ = -64.
Similarly, a × a × a × b × b can be expressed as a³b² (read as a cubed b squared).
a × a × b × b × b × b can be expressed as a²b⁴ (read as a squared b raised to the power 4).
Remember that 4 + 4 + 4 = 3 × 4 = 12, whereas 4 × 4 × 4 = 4³ = 64.

Express the number 32400 as a product of its prime factors and represent the prime factors in their exponential form.

32400 = 2 × 2 × 2 × 2 × 5 × 5 × 3 × 3 × 3 × 3.
In exponential form, this would be 32400 = 2⁴ × 5² × 3⁴.

The Stones that Shine…
Three daughters with curious eyes,
Each got three baskets, a kingly prize.
Each basket had three silver keys,
Each opens three big rooms with ease.
Each room had tables one, two, three,
With three bright necklaces on each, you see.
Each necklace had three diamonds so fine…
Can you count these stones that shine?

Find out the number of baskets and rooms. The information given can be visualised as shown below.

From the diagram, the number of rooms is 34. This can be computed by repeatedly multiplying 3 by itself,
3 × 3 = 9.
9 × 3 = 27.
27 × 3 = 81.
81 × 3 = 243.

How many diamonds were there in total? Can we find out by just one multiplication using the products above?
The number of diamonds is 3 × 3 × 3 × 3 × 3 × 3 × 3 = 3⁷.
We can write 3⁷ = (3 × 3 × 3 × 3) × (3 × 3 × 3)

We had computed till 3⁴.
To find 3⁷, we can just multiply 3⁴ (= 81) by 3³ (= 27).
= 3⁴ × 3³
= 81 × 27
= 2187

3⁷ can also be written as 3² × 3⁵. Can you reason out why?
This can be easily extended to products where exponents are the same letter-numbers.

Write the product p⁴ × p⁶ in exponential form.
p⁴ × p⁶ = (p × p × p × p) × (p × p × p × p × p × p) = p¹⁰
We can generalise this to n^a × n^b = n^a+b, where a and b are counting numbers.

Use this observation to compute the following.
(i) 4⁶
4⁶ can be evaluated in these two ways,

Similarly, 7⁴ = (7 × 7) × (7 × 7) = 7² × 7² = (7²)², and
2¹⁰ = (2 × 2) × (2 × 2) × (2 × 2) × (2 × 2) × (2 × 2)
= (2²) × (2²) × (2²) × (2²) × (2²)
= (2²)⁵

Is 2¹⁰ also equal to (2⁵)²? Write it as a product.
2¹⁰ = (2 × 2 × 2 × 2 × 2) × (2 × 2 × 2 × 2 × 2)
= (2⁵) × (2⁵)
= (2⁵)²

In general, (n^a)^b = (n^b)^a = n^a×b = n^ab, where a and b are counting numbers.

Magical Pond
In the middle of a beautiful, magical pond lies a bright pink lotus. The number of lotuses doubles every day in this pond. After 30 days, the pond is completely covered with lotuses. On which day was the pond half full?

If the pond is completely covered by lotuses on the 30th day, how much of it is covered by lotuses on the 29th day? Since the number of lotuses doubles every day, the pond should be half covered on the 29th day. There is another pond in which the number of lotuses triples every day. When both the ponds had no flowers, Damayanti placed a lotus in the doubling pond. After 4 days, she took all the lotuses from there and put them in the tripling pond. How many lotuses will be in the tripling pond after 4 more days?
After the first 4 days, the number of lotuses is 1 × 2 × 2 × 2 × 2 = 2⁴.
After the next 4 days, the number of lotuses is 2⁴ × 3 × 3 × 3 × 3 = 2⁴ × 3⁴.

What if Damayanti had changed the order in which she placed the flowers in the lakes? How many lotuses would be there?
1 × 3⁴ × 2⁴ = (3 × 3 × 3 × 3) × (2 × 2 × 2 × 2).

Can this product be expressed as an exponent mn, where m and n are some counting numbers?
By regrouping the numbers (3 × 2) × (3 × 2) × (3 × 2) × (3 × 2)
= (3 × 2)⁴
= 6⁴

In general form, m^a × n^a = (mn)^a, where a is a counting number.
In general, we can show that \(\frac{m^a}{n^a}=\left(\frac{m}{n}\right)^a\)

How Many Combinations
Estu has 4 dresses and 3 caps. How many different ways can Estu combine the dresses and caps?
For each cap, he can choose any of the 4 dresses, so for 3 caps, 4 + 4 + 4 = 4 × 3 = 12 combinations are possible.
We can also look at it as for each dress, Estu can choose any of the 3 caps, so for 4 outfits, 3 + 3 + 3 + 3 = 3 × 4 = 12 combinations are possible.

Roxie has 7 dresses, 2 hats, and 3 pairs of shoes. How many different ways can Roxie dress up?
Try drawing a diagram like the one above.

Estu and Roxie came across a safe containing old stamps and coins that their great-grandfather had collected. It was secured with a 5-digit password. Since nobody knew the password, they had no option except to try every password until it opened. They were unlucky, and the lock only opened with the last password, after they had tried all possible combinations. How many passwords did they end up checking?

Instead of a 5-digit lock, let us assume we have a 2-digit lock and try to find out how many passwords are possible. There are 10 options for the first digit (0 to 9). For each of these, there are 10 options for the second digit (If 0 is the first digit, then 00, 01, 02, 03, …, 09 are possible). Therefore, the total number of combinations for a 2-digit lock is 10 × 10 = 100.

Now, suppose we have a 3-digit lock. For each of the earlier 100 (2-digit) passwords, there are 10 choices for the third digit. So, there are 100 × 10 = 1000 combinations for a 3-digit lock. You can list them all: 000, 001, 002,…, 997, 997, 999.

How many 5-digit passwords are possible?

Each digit has 10 choices, so a 5-digit lock will have 10 × 10 × 10 × 10 × 10 = 105 = 1,00,000 passwords.
This is the same as writing numbers till 99,999 with all 5 digits, i.e., 00000, 00001, 00002, …00010, 00011, …, 00100, 00101, …, 00999, …, 30456, …, 99998, 99999.
Estu says, “Next time, I will buy a lock that has 6 slots with the letters A to Z. I feel it is safer.”

How many passwords are possible with such a lock? Think about how many combinations are possible in different contexts. Some examples are

• Pincodes of places in India – The Pincode of Vidisha in Madhya Pradesh is 464001. The Pincode of Zemabawk in Mizoram is 796017.
• Mobile numbers.
• Vehicle registration numbers.

Try to find out how these numbers or codes are allotted/generated.

The Other Side of Powers Class 8 Notes

Imagine a line of length 16 units.
Erasing half of it would result in 2⁴ ÷ 2 = \(\frac{2 \times 2 \times 2 \times 2}{2}\)
= 2 × 2 × 2
= 2³
= 8 units.
Erasing half one more time would result in (2⁴ ÷ 2) ÷ 2 = 2⁴ ÷ 2²
= \(\frac{2 \times 2 \times 2 \times 2}{2 \times 2}\)
= 2 × 2
= 2²
= 4 units.
Halving 16 cm three times may be written as 2⁴ ÷ 2³ = \(\frac{2 \times 2 \times 2 \times 2}{2 \times 2 \times 2}\)
= 2
= 2¹
= 2 units.
From this, we can see that 2⁴ ÷ 2³ = 2^4-3 = 2¹.
What is 2¹⁰⁰ ÷ 2²⁵ in powers of 2?
In a generalised form, n^a ÷ n^b = n^a-b,
where n ≠ 0 and a, and b are counting numbers and a > b.

We have not covered the case when the exponent is 0; for example, what is 20?
Let us define 20 in a way that the generalised form above holds.
\(2^0=2^{4-4}=2^4 \div 2^4=\frac{2 \times 2 \times 2 \times 2}{2 \times 2 \times 2 \times 2}=1\)
In fact, for any letter number a
2⁰ = 2^a-a = 2^a ÷ 2^a = 1.
In general, x^a ÷ x^a = x^a-a = x⁰,
and so 1 = x⁰, where x ≠ 0 and a, a counting numbers.

When Zero is in Power!

When a line of length 24 units is halved 5 times,
2⁴ ÷ 2⁵ = \(\frac{2 \times 2 \times 2 \times 2}{2 \times 2 \times 2 \times 2 \times 2}=\frac{1}{2}\) units.
Using the generalised form, we get 2⁴ ÷ 2⁵ = 2^(4-5) = 2^-1.
So, 2^-1 = \(\frac{1}{2}\)
When a line of length 24 units is halved 10 times,
we get 2⁴ ÷ 2¹⁰ = 2^(4-10) = 2^-6 units.
When expanded, 2⁴ ÷ 2¹⁰ = \(\frac{2 \times 2 \times 2 \times 2}{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2}=\frac{1}{2^6}=\frac{1}{64}\), which is also written as 2^-6.
Similarly, 10^-3 = \(\frac{1}{10^3}\), 7-2 = \(\frac{1}{7^2}\), etc.

Can we write 10³ = \(\frac{1}{10^{-3}}\)?
We can write \(\frac{1}{10^{-3}}=\frac{1}{1 / 10^3}=1 \div \frac{1}{10^3}\) = 1 × 10³ = 10³
Similarly, 7² = \(\frac{1}{7^{-2}}\) and 4^a = \(\frac{1}{4^{-a}}\)
In a generalised form, n^-a = \(\frac{1}{n^a}\) and n^a = \(\frac{1}{n^{-a}}\), where n ≠ 0.

Consider the following general forms we have identified.
n^a × n^b = n^a+b
(n^a)^b = (n^b)^a = n^a×b
n^a ÷ n^b = n^a-b

Power Lines
Let us arrange the powers of 4 along a line.

Powers of 10 Class 8 Notes

We have used numbers like 10, 100, 1000, and so on when writing Indian numerals in an expanded form.
For example, 47561 = (4 × 10000) + (7 × 1000) + (5 × 100) + (6 × 10) + 1.
This can be written using powers of 10 as (4 × 10⁴) + (7 × 10³) + (5 × 10²) + (6 × 10¹) + (1 × 10⁰).

How can we write 561.903?
561.903 = (5 × 100) + (6 × 10) + 1 + (9 × \(\frac{1}{10}\)) + (0 × \(\frac{1}{100}\)) + (3 × \(\frac{1}{1000}\)).
Writing it using powers of 10, we have
561.903 = (5 × 10²) + (6 × 10¹) + (1 × 10⁰) + (9 × 10^-1) + (0 × 10^-2) + (3 × 10^-3).

Scientific Notation
Let’s look at some facts involving large numbers

• The Sun is located 30,00,00,00,00,00,00,00,00,000 m from the centre of our Milky Way galaxy.
• The number of stars in our galaxy is 1,00,00,00,00,000.
• The mass of the Earth is 59,76,00,00,00,00,00,00,00,00,00,000 kg.

As the number of digits increases, it becomes difficult to read the numbers correctly. We may miscount the number of zeroes or place commas incorrectly. We will then read the wrong value. It is like getting ₹5,000 when you were supposed to get ₹50,000. The number of zeroes is more important than the initial digits in several cases.

Can we use the exponential notation to simplify and read these very large numbers correctly?
For example, the number 5900 can be expressed as
5900 = 590 × 10 = 590 × 10¹
= 59 × 100 = 59 × 10²
= 5.9 × 1000 = 5.9 × 10³
= 0.59 × 10000 = 0.59 × 10⁴
Any number can be written as the product of a number between 1 and 10 and a power of 10.
For example, 5900 = 5.9 × 10³
20800 = 2.08 × 10⁴
80,00,000 = 8 × 10⁶

Write the large-number facts we read just before in this form. In scientific notation or scientific form (also called standard form), we write numbers as x × 10^y, where x ≥ 1 and x < 10 is the coefficient and y, the exponent, is any integer. Often, the exponent y is more important than the coefficient x. When we write the 2 crore population of Mumbai as 2 × 10⁷, the 7 is more important than the 2. Indeed, if the 2 is changed to 3, the population increases by one-half, i.e., 2 crore to 3 crore, whereas if the 7 is changed to 8, the population change is 10 times, i.e., 2 crores to 20 crores. Therefore, the standard form explicitly mentions the exponent, which indicates the number of digits.

If we say that the population of Kohima is 1,42,395, then it gives the impression that we are quite sure about this number up to the unit’s place. When we use large numbers, in most cases, we are more concerned about how big a quantity or measure is, rather than the exact value. If we are only sure that the population is around 1 lakh 42 thousand, we can write it as 1.42 × 10⁵. If we can only be certain that it is around 1 lakh 40 thousand, we write it as 1.4 × 10⁵. The number of digits in the coefficient reflects how well we know the number. The most important part of any number written in scientific form is the exponent, and then the first digit of the coefficient. The digits following the coefficient are small corrections to the first digit.

These values are rounded-off estimates, averages, or approximations; most of the time, they serve the purpose at hand.
The distance between the Sun and Saturn is 14,33,50,00,00,000 m = 1.4335 × 10¹² m.
The distance between Saturn and Uranus is 14,39,00,00,00,000 m = 1.439 × 10¹² m.
The distance between the Sun and Earth is 1,49,60,00,00,000 m = 1.496 × 10¹¹ m.

Can you say which of the three distances is the smallest?

The number line below shows the distance between the Sun and Saturn (1.4335 × 10¹² m). On the number line below, mark the relative position of the Earth. The distance between the Sun and the Earth is 1.496 × 10¹¹ m.

Did You Ever Wonder? Class 8 Notes

Last year, we looked at interesting thought experiments in the chapter on Large Numbers. Let us continue this journey. Nanjundappa wants to donate jaggery equal to Roxie’s weight and wheat equal to Estu’s weight. He is wondering how much it would cost.

What would be the worth (in rupees) of the donated jaggery? What would be the worth (in rupees) of the donated wheat?
In order to find out, let us first describe the relationships among the quantities present.
Worth of jaggery (in rupees) = Roxie’s weight in kg × cost of 1 kg of jaggery.
Worth of wheat (in rupees) = Estu’s weight in kg × cost of 1 kg of wheat.

Make necessary and reasonable assumptions for the unknowns and find the answers. Remember, Roxie is 13 years old and Estu is 11 years old.
Assuming Roxie’s weight to be 45 kg and the cost of 1 kg of jaggery to be ₹70, the worth of donated jaggery is 45 × 70 = ₹3150.
Assuming Estu’s weight to be 50 kg and the cost of 1 kg of wheat to be ₹50, the worth of donated wheat is 50 × 50 = ₹2500.

The practice of offering goods equal to the weight of a person, called Tulabhara or Tulabharam, is quite old and is still followed in many places in Southern India. It is a symbol of bhakti (surrendering oneself), a token of gratitude; it also supports the community.

Initially, your guesses may be very far off from the answer, and it is perfectly fine! You will get better at it as you do it often and in different situations. Guessing and estimating can build intuition about numbers and various quantities.

Estu asks, “What if we use 5-rupee coins or 10-rupee notes instead? How much money could it be?” Make an instinctive guess first. Then find out (make necessary and reasonable assumptions about the unknown details and find the answers).

Estu says, “When I become an adult, I would like to donate notebooks worth my weight every year”. Roxie says, “When I grow up, I would like to do annadana (offering grains or meals) worth my weight every year”.

How many people might benefit from each of these offerings in a year? Again, guess first before finding out.
Roxie and Estu overheard someone saying, “We did a padayatra for about 400 km to reach this place! We arrived early this morning.”

Before the rise of modern transport, people moved from one place to another by walking. Sometimes merchants, sages, and scholars walked thousands of kilometres to different parts of the world across deserts, mountains, and rivers.

Linear Growth vs. Exponential Growth
Roxie tells Estu about a science fiction novel she is reading where they build a ladder to reach the moon, “… I wonder if we had a ladder like that, how many steps would it have?”.

What do you think? Make an instinctive guess first. Would the number of steps be in thousands, lakhs, crores, or even more?
To find out, we would need to know the gap between consecutive steps of the ladder. Let’s assume a reasonable distance of 20 cm. Visualising the problem as shown.

We have to find out how many 20 cm make 3,84,400 km. If we calculate the value, we get the result as 1,92,20,00,000 steps, which is 192 crore and 20 lakh steps or 1 billion 922 million steps. The fixed increase in the distance from the Earth with each step (a 20 cm gain after each step) is called linear growth. To cover the distance between the Earth and the Moon, it takes 1,92,20,00,000 steps with linear growth, whereas it takes just 46 folds of a piece of paper with exponential growth! Linear growth is additive, whereas exponential growth is multiplicative.

Some examples of exponential growth we have seen earlier in this chapter are ‘The Stones that Shine’, ‘Magical Pond’, and ‘How Many Combinations’. We shall explore more such interesting examples in a later chapter and also in the next grade.

Getting a Sense for Large Numbers
Last year, we learnt about lakhs and crores, as well as millions and billions. A lakh is 105 (1,00,000), a crore is 107 (1,00,00,000), and an arab is 109 (1,00,00,00,000), whereas a million is 106 (1,000,000) and a billion is 109 (1,000,000,000).

You might know the size of the world’s human population. Have you ever wondered how many ants there might be in the world or how long ago humans emerged? In this section, we shall explore numbers significantly larger than arabs and billions. We shall use powers of 10 to represent and compare these numbers in each case.

10⁰ – As of mid-2025, only two northern white rhinos are remaining in the world, both females, and they reside at the Ol Pejeta Conservancy in Kenya (= 2 × 10⁰).

10¹ – As of early 2024, the total population of Hainan gibbons is a meagre 42 (≈ 4 × 10¹).

10² – There are just 242 Kakapo alive as of mid-2025 (≈ 2 × 10²).

10³ – There are fewer than 3000 Komodo dragons in the world, all based in Indonesia (≈ 3 × 10³).

10⁴ – A 2005 estimate of the maned wolf population showed that there are more than 17000 of them; most are located in Brazil (1.7 × 10⁴).

10⁵ – As of 2018, there are around 4.15 lakh African elephants (≈ 4 × 10⁵).

10⁶ – There are an estimated 50 lakh/5 million American alligators as of 2025 (5 × 10⁶).

10⁷ – The global camel population is estimated to be over 3.5 crore/35 million (3.5 × 10⁷). India has only about 2.5 lakhs of them. The global horse population is around 5.8 crore/58 million (5.8 × 10⁷), with about half of them in America.

10⁸ – More than 20 crore/200 million (2 × 10⁸) water buffaloes are estimated worldwide, with a vast majority of them in Asia.

10⁹ – The estimated global population of starlings is around 1.3 arab/1.3 billion. The global human population as of 2025 is 8.2 arab/8.2 billion (8.2 × 10⁹).

A picture of a starling murmuration over a farm in the UK. Starling murmuration is a mesmerising aerial display of thousands of starlings flying in synchronised, swirling patterns. It is often described as a ‘choreographed dance’.

With a global human population of about 8 × 10⁹ and about 4 × 10⁵ African elephants, can we say that there are nearly 20,000 people for every African elephant?
10¹⁰ – The global chicken population living at any time is estimated at ≈ 33 billion (3.3 × 10¹⁰).

10¹² – The estimated number of trees (2023) globally stands at 30 kharab/3 trillion (3 × 10¹²). One kharab is 100 arab, and one trillion is 1000 billion.

10¹⁴ – The estimated mosquito population worldwide (2023) is 11 neel/110 trillion. A derived estimate of the population of the Antarctic krill stands at 50 neel/500 trillion (5 × 10¹⁴).

10¹⁵ – An estimate of the beetle population stands at 1 padma/1 quadrillion (1 × 10¹⁵). The estimate of the earthworm population is also at 1 padma/1 quadrillion.

10¹⁶ – The estimated population of ants globally is 20 padma/20 quadrillion (2 × 10¹⁶). Ants alone outweigh all wild birds and wild mammals combined.

10²¹ is supposed to be the number of grains of sand on all beaches and deserts on Earth. This is enough sand to give every ant 10 little sand castles to live in.

10²³ – The estimated number of stars in the observable universe is 2 × 10²³.

10²⁵ – There are an estimated 2 × 10²⁵ drops of water on Earth (assuming 16 drops per millilitre).

A different way to say your age!
“How old are you?” asked Estu.
“I completed 13 years a few weeks ago!” said Roxie.
“How old are you?” asked Estu again.
“I’m 4840 days old today!” said Roxie.
“How old are you?” asked Estu again.
“I’m 116160 hours old!” said Roxie.
Estimate finding this number.
Estu: “I am 4070 days old today. Can you find out my date of birth?”
If you have lived for a million seconds, how old would you be?
We shall look at approximate times and timelines of some events and phenomena, and use powers of 10 to represent and compare these quantities.

Time in Seconds – Comparison to real-world events/phenomena
10⁰ = 1 second – Time taken for a ball thrown up to fall back on the ground (typically a few seconds).

10¹ = 10 seconds – Time blood takes to complete one full circulation through the body: 10 – 20 seconds (1 × 10¹ – 2 × 10¹ seconds). Typical waiting time at a traffic signal.

10² seconds ≈ 1.6 minutes – Time needed to make a cup of tea: 5 – 10 minutes (≈ 4 × 10² – 8 × 10² seconds). Time for light to reach the Earth from the Sun: about 8 minutes (≈ 5 × 10² seconds).

10³ seconds ≈ 16.6 minutes – Satellites in low Earth orbits take between 90 minutes (≈ 5.5 × 10³ seconds) to 2 hours to complete one full revolution around the Earth.

10⁴ seconds ≈ 2.7 hours – The time needed to digest a meal: about 2 – 4 hours to pass through the stomach. Lifespan of an adult mayfly: about a day (≈ 9 × 10⁴ seconds).

10⁵ seconds ≈ 1.16 days, and 10⁶ seconds ≈ 11.57 days.

10⁷ seconds ≈ 115.7 days/≈ 3.8 months – Time spent sleeping in a year: about 4 months. Time taken by Mangalyaan mission to reach Mars: 298 days (≈ 2.65 × 10⁷ seconds). Time taken by Mars for one full revolution around the Sun: 687 Earth-days/1.88 Earth-years (≈ 6 × 10⁷ seconds).

10⁸ seconds ≈ 3.17 years – The typical lifespan of most dogs is 3 to 15 years.

10⁹ seconds ≈ 31.7 years – The orbital period of Halley’s comet is 75 – 79 years; the next expected return is in the year 2061 (≈ 2.4 × 10⁹ seconds). Duration of one full revolution of Neptune around the Sun: 60,190 Earth-days/~165 Earth-years or 89,666 Neptunian days/1 Neptunian-year (≈ 5.2 × 10⁹ seconds). A day on Neptune is about 16.1 hours

Notice how rapid exponential growth is 106 seconds is less than a fortnight, but 109 seconds is a whopping 31 years (about half the life expectancy of a human)!

10¹⁰ seconds ≈ 317 years – The Chola dynasty ruled for more than 900 years (≈ 3 × 10¹⁰ seconds) between the 3rd Century BCE and 12th Century CE.

10¹¹ seconds ≈ 3,170 years – Age of the oldest known living tree: about 5000 years (≈ 1.57 × 10¹¹ seconds). Time since the last peak ice age: 19,000 – 26,000 years ago (≈ 6 × 10¹¹ seconds – 8.2 × 10¹¹ seconds).

10¹² seconds ≈ 31,700 years – Early Homo sapiens first appeared 2-3 lakh years ago (≈7 × 10¹² – 9 × 10¹² seconds). The entire population around that time could fit in a large cricket stadium.

10¹³ seconds ≈ 3.17 lakh years – The Steppe Mammoth is estimated to have appeared around 8 – 18 lakh years ago.

10¹⁴ seconds ≈ 3.17 million years – A fossil of Kelenken Guillermoi, a type of terror bird, is dated to 15 million years ago.

10¹⁵ seconds ≈ 3.17 crore years – Age of Himalayas: 5.5 crore years/55 million years (≈ 1.7 × 10¹⁵ seconds); they continue to grow a few mm every year. Dinosaurs went extinct 6.6 crore years ago/66 million years ago (≈ 2 × 10¹⁵ seconds). Dinosaurs first appeared more than 20 crore/200 million years ago (≈ 6 × 10¹⁵ seconds). It takes about 23 crore years for the Sun to make one complete trip around the Milky Way (≈ 7 × 10¹⁵ seconds).

10¹⁶ seconds ≈ 31.7 crore years – Plants on land started 47 crore/470 million years ago.

10¹⁷ seconds ≈ 3.17 billion years – The oldest fossil evidence suggests that bacteria first appeared about 3.7 billion years ago. The Earth is 4.5 billion years old. The Milky Way galaxy was formed 13.6 billion years ago, and the Universe was formed 13.8 billion years ago.

Notice that 10⁹ seconds is of the order of the lifespan of a human, whereas 10¹⁸ seconds ago the universe did not exist according to modern physics!! The exponential notation can concisely capture very large quantities.

A Pinch of History Class 8 Notes

In the Lalitavistara, a Buddhist treatise from the first century BCE, we see number names for odd powers of ten up to 10⁵³. The following occurs as part of the dialogue between the mathematician Arjuna and Prince Gautama, the Bodhisattva. “Hundred kotis are called an ayuta (10⁹), hundred ayutas a niyuta (10¹¹), hundred niyutas a kankara (10¹³), …, hundred sarva-balas a visamjnagati (10⁴⁷), hundred visamjna-gatis a sarvajna (10⁴⁹), hundred sarvajnas a vibhutangama (10⁵¹), a hundred vibhutangamas is a tallakshana (10⁵³).” Mahaviracharya gives a list of 24 terms (i.e., up to 10²³) in his treatise Ganita-sara-sangraha. An anonymous Jaina treatise, Amalasiddhi, gives a list with a name for each power of ten up to 10⁹⁶ (dasha-ananta). A Pali grammar treatise of Kaccayana lists number-names up to 10¹⁴⁰, named asankhyeya.

For expressing high powers of ten, Jaina and Buddhist texts use bases like sahassa (thousand) and koti (ten million); for instance, prayuta (10⁶) would be dasa sata sahassa (ten hundred thousand). The modern naming is similar to this, where we say,
A hundred thousand is a lakh
100 × 1000 = 1,00,000
10² × 10³ = 10⁵

A hundred lakhs is a crore
100 × 1,00,000 = 1,00,00,000
10² × 10⁵ = 10⁷

A hundred crores is an arab
100 × 1,00,00,000 = 1,00,00,00,000
10² × 10⁷ = 10⁹

A hundred Arabs is a kharab
100 × 1,00,00,00,000 = 1,00,00,00,00,000
10² × 10⁹ = 10¹¹

Continuing this, a hundred kharab is a neel (10¹³), a hundred neel is a padma (10¹⁵), a hundred padma is a shankh (10¹⁷), and a hundred shankh is a maha shankh (10¹⁹).
In the American/International system, we say

A thousand thousand is a million
1000 × 1000 = 1,000,000
10³ × 10³ = 10⁶

A thousand million is a billion
1000 × 1,000,000 = 1,000,000,000
10³ × 10⁶ = 10⁹

A thousand billion is a trillion
1000 × 1,000,000,000 = 1,00,000,000,000
10³ × 10⁹ = 10¹²

Continuing this, a thousand trillion is a quadrillion (10¹⁵). This pattern continues. Observe the names million (10⁶), billion (10⁹), trillion (10¹²), quadrillion (10¹⁵), quintillion (10¹⁸), sextillion (10²¹), septillion (10²⁴), octillion (10²⁷), nonillion (10³⁰), decillion (10³³).

What does the first part of each name denote?
The number 10100 is also called a googol. The estimated number of atoms in the universe is 10⁷⁸ to 10⁸². The number 10googol is called a googolplex. It is hard to imagine how large this number is!

The currency note with the highest denomination in India currently is 2000 rupees. Guess what is the highest denomination of a currency note ever, across the world? The highest numerical value banknote ever printed was a special note valued 1 sextillion pengo (1021 or 1 milliard bilpengo) printed in Hungary in 1946, but it was never issued. In 2009, Zimbabwe printed a 100 trillion (10¹⁴) Zimbabwean dollar note, which at the time of printing was worth about $30.

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