Number Play Class 8 Notes Maths Chapter 5 - #NCSOLVE 📚

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Class 8 Maths Chapter 5 Notes Number Play

Class 8 Maths Notes Chapter 5 – Class 8 Number Play Notes

→ We explored and learnt various properties of divisibility:

• If a is divisible by b, then all multiples of a are divisible by b.
• If a is divisible by b, then a is divisible by all the factors of b.
• If a divides m and a divides n, then a divides m + n and m-n.
• If a is divisible by b and is also divisible by c, then a is divisible by the LCM of b and c.

→ We learnt shortcuts to check divisibility by 3, 9, and 11, and why they work.

→ Through all this, we were exposed to the power of mathematical thinking and reasoning, using algebra, visualisation, examples, and counterexamples.

Is this a Multiple of?

If a number ‘a’ divides another number ‘b’ exactly, then we say that ‘a’ is a factor of ‘b’ and ‘b’ is a multiple of ‘a’.

Factor: A factor of a number is an exact divisor of that number.

Multiple: A number is said to be a multiple of any of its factors.
Example: When 20 is divided by 5, the remainder is zero, i.e., 5 divides 20 exactly. So 5 is a factor of 20 and 20 is a multiple of 5.

Even numbers: All the multiples of 2 are called even numbers.
Example: 2, 4, 6, 8, 10, 12,… are all even numbers.

Odd numbers: Numbers that are not multiples of 2 are called odd numbers.
Example: 1, 3, 5, 7, 9, 11, 13, 15, … are all odd numbers.

Sum of Consecutive Numbers

Consecutive Numbers:
Consecutive numbers are integers that follow each other in order, with a fixed difference of 1 between each pair.
For example, 4, 5, 6, 7 are consecutive numbers because each number increases by 1.

• Consecutive numbers can also be negative or include zero.
• For instance, -3, -2, -1, 0, 1 are also consecutive numbers. They form an arithmetic sequence with a common difference of 1.
• If ‘x’ represents the first number, then the consecutive numbers can be defined as: x, x + 1, x + 2, and so on.
• For example, if the first number is 7, the consecutive numbers are 7, 7 + 1 = 8, and 7 + 2 = 9.

Properties of the Sum of Consecutive Numbers

• Sum of two consecutive numbers: Always an odd number.
• Sum of two consecutive odd numbers: Always an even number.
• Sum of two consecutive even numbers: Always an even number.
• The sum of ‘n’ consecutive numbers is divisible by if is an odd number.
• For instance, the sum of three consecutive numbers is always divisible by 3.

Solving problems involving sums of consecutive numbers
When the sum is given, we can set up an algebraic equation based on the representation of consecutive numbers (e.g., x, x + 1, x + 2) and solve for ‘x’.

Breaking Even
To determine whether an arithmetic expression is even without computing its exact value, we can use parity rules.
Parity Rules: The sum or difference of two numbers has an even parity (is even) if both numbers have the same parity (both even or both odd), and odd parity (is odd) if the numbers have different parities (one even and one odd).
Examples:

• Even + Even = Even: When you add two even numbers (e.g., 2 + 4 = 6), the result is always even.
• Odd + Odd = Even: When you add two odd numbers (e.g., 3 + 5 = 8), the result is also even.
• Even + Odd = Odd: When you add an even and an odd number (e.g., 2 + 3 = 5), the result is always odd.
• Even – Even = Even: Subtracting an even number from another even number (e.g., 6 – 2 = 4) results in an even number.
• Odd – Odd = Even: Subtracting an odd number from another odd number (e.g., 7 – 3 = 4) results in an even number.
• Even – Odd = Odd: Subtracting an odd number from an even number (e.g., 6 – 3 = 3) or vice versa results in an odd number.

Always, Sometimes, or Never
To determine if a statement about factors and multiples is “Always True,” “Sometimes True,” or “Never True,” we need to analyze its validity across different scenarios.

• “Always True” means it holds for all possible cases.
• “Sometimes True” means it is true for some cases but not all.
• “Never True” means it is false in every instance.

If a number a divides both M and N, then it will also divide their sum (M + N) and their difference (M – N). This is a fundamental property of divisibility.

If a number ‘A’ is divisible by another number ‘k’, then any multiple of ‘A’ will also be divisible by ‘k’.

If a number A is divisible by both k and m, then it will be divisible by the least common multiple (LCM) of k and m.

Numbers are often written in the form of edcba, each letter representing a digit at a place value:

• e × 10000 (ten-thousands)
• d × 1000 (thousands)
• c × 100 (hundreds)
• b × 10 (tens)
• a × 1 (units)

Checking Divisibility Quickly

Tests of Divisibility by 10
A number is divisible by 10 if its unit digit is zero.
Justification: Let us consider a four-digit number n = abcd where a ≠ 0 is the thousandth digit and can take values from 1 to 9, and b, c, and d are the hundred, ten, and unit digits, each of which can take values from 0 to 9.
In expanded form, n = abcd
= 1000 × a + 100 × b + 10 × c + d
= 10[100a + 10b + c] + d
= 10k + d, where k = 100a + 10b + c
∴ n = 10k + d ……(1)
where d is the unit digit.
If d = 0, then n = 10k, which is divisible by 10.
∴ A number is divisible by 10 if its unit digit d is 0.
If d ≠ 0, then from (1), n-d = 10k, which is divisible by 10.
Hence, we learn that if we subtract the digit at the ones place from any number, it becomes divisible by 10.
Clearly, the above justification of the test of divisibility by 10 holds also for two-digit, three-digit numbers also and so on.

Test of Divisibility by 5
If the sum of the digits of a natural number is divisible by 5, then the number is divisible by 5.
A number is divisible by 5 if its unit digit is 0 or 5.
Justification: Reproduce Eqn. (1) to get n = 10k + d
If d = 0, then n = 10k = 5 × 2k, which is divisible by 5.
If d = 5; then n = 10k + 5 = 5(2k + 1) which is divisible by 5.
∴ A number is divisible by 5 if its unit place is 0 or 5.
From equation (1), n = 10k + d.
Since 10k is divisible by 5 therefore when a natural number n is divided by 5, the remainder will be equal to the remainder when ones digit d is divided by 5.
For example, if 237 is divided by 5, then the remainder = the remainder obtained on dividing 7 by 5 = 2.
Again, if 623 is divided by 5, then the remainder = remainder on dividing 3 by 5 = 3.

Test of Divisibility by 2
A number is divisible by 2 if its unit digit is an even digit, i.e., 0, 2, 4, 6, 8.
Justification: Reproduce equation (1) to get n = 10k + d
We know that n will be divisible by 2 if n is even.
Also 10k = 2(5k) is even.
∴ d must also be even because the sum n = 10k + d of an even number 10k.
∴ Unit digit d = 0, 2, 4, 6, 8.
In all the above three tests of divisibility by 10, 5, and 2, the divisibility is decided just by the unit digit.

Now we shall learn two more tests of divisibility (by 3 and by 9) in which all the digits of the given number will be used to test the divisibility.
Justification: Let n = abc be a three-digit number.
∴ n = 100 × a + 10 × b + c
= 100a + 10b + c
= 99a + a + 9b + b + c
= 99a + 9b + a + b + c
= 9(11a + b) + (a + b + c) ……..(2)
Now 9(11a + b) is a multiple of 9.
Therefore, from (1), n is a multiple of 9 if a + b + c is a multiple of 9.
i.e., a number n = abc is divisible by 9 if the sum of the digits of the number, i.e., (a + b + c), is divisible by 9.
We can claim from equation (2) above, that when a number is divided by 9, the remainder is equal to the remainder when the sum of its digits (a + b + c) is divided by 9. [∵ 9(11a + b) leaves remainder 0 when divided by 9]

Divisibility by 6
Justification: We have 6 = 2 × 3, where 2 and 3 have no common factor.
∴ A number is divisible by 6 if it is divisible by 2 and 3.
Hence, if a number is not divisible by 2 or by 3 or by both, then it is not divisible by 6.

Divisibility by 11
A number is divisible by 11 if the difference of the sum of its digits in odd places and the sum of its digits in even places is either 0 or a multiple of 11, i.e., divisible by 11.
Justification: Let n = abcd be a four-digit number.
In expanded form, n = abcd
= 1000 × a + 100 × b + 10 × c + d
= 1000a + 100b + 10c + d
= 1001a – a + 99b + b + 11c – c + d
= 1001a + 99b + 11c + b + d – a – c
= 11(91a + 9b + c) + (b + d) – (a + c)
∴ n = abcd = Multiple of 11 + (b + d) – (a + c) …..(iv)
∴ n is divisible by 11 if either (b + d) – (a + c) = 0 or is a multiple of 11. i.e., divisible by 11.
(b + d) → sum of digits in even places
(a + c) → sum of digits in odd places

Digital Roots
The digital root of a number is the single-digit value obtained by repeatedly summing the digits of the number until only one digit remains.
Example 1: Number = 999999
Step 1: 9 + 9 + 9 + 9 + 9 + 9 = 54
Step 2: 5 + 4 = 9
Digital root = 9

Example 2: Number = 2024
Step 1: 2 + 0 + 2 + 4 = 8
Digital root = 8

Digits in Disguise
Cryptarithms are puzzles in which letters take the place of digits, and the problem is to find which letter represents which digit.
In this section, we shall discuss problems on addition and multiplication only. Here are two rules we follow while doing such puzzles.

• Each letter in the puzzle must stand for just one digit.
• Each digit must be represented by just one letter.
• The first digit of a number cannot be zero.
• Thus, we write the number ‘fifty-eight’ as 58 and not as 058 or 0058.

Is This a Multiple Of? Class 8 Notes

Sum of Consecutive Numbers
Anshu is exploring the sums of consecutive numbers. He has written the following.

Now, he is wondering

• “Can I write every natural number as a sum of consecutive numbers?”
• “Which numbers can I write as the sum of consecutive numbers in more than one way?”
• “Ohh, I know all odd numbers can be written as a sum of two consecutive numbers. Can we write all even numbers as a sum of consecutive numbers?”
• “Can I write 0 as a sum of consecutive numbers? Maybe I should use negative numbers.”

Explore these questions and any others that may occur to you. Discuss them with the class.
Take any 4 consecutive numbers. For example, 3, 4, 5, and 6. Place ‘+’ and ‘–’ signs in between the numbers. How many different possibilities exist? Write all of them.

Eight such expressions are possible. You can use the diagram below to systematically list all the possibilities.

Evaluate each expression and write the result next to it. Do you notice anything interesting?
Now, take four other consecutive numbers. Place the ‘+’ and ‘–’ signs as you have done before. Find out the results of each expression. What do you observe?
Repeat this for one more set of 4 consecutive numbers. Share your findings

Some sums always appear, no matter which 4 consecutive numbers are chosen. Isn’t that interesting?
Do these patterns occur no matter which 4 consecutive numbers are chosen? Is there a way to find out through reasoning?
Hint: Use algebra and describe the 8 expressions in a general form.

You might have noticed that the results of all expressions are even numbers. Even numbers have a factor of 2. Negative numbers having a factor of 2 are also even numbers, for example, -2, -4, -6, and so on. Check if anyone in your class got an odd number. When 4 consecutive numbers are chosen, no matter how the ‘+’ and ‘–’ signs are placed between them, the resulting expressions always have even parity.

Now take any 4 numbers, place ‘+’ and ‘–’ signs in the eight different ways, and evaluate the resulting expression. What do you observe about their parties? Repeat this with other sets of 4 numbers.

Is there a way to explain why this happens?
Hint: Think of the rules for parity of the sum or difference of two numbers.

Explanation 1: Let us consider any of the 8 expressions formed by four numbers a, b, c, and d.
When one of its signs is switched, its value always increases or decreases by an even number!
Let us see why.
Consider one of the expressions: a + b – c – d.
Replacing +b by -b, we get a – b – c – d.
By how much has the number changed?
It has changed by (a + b – c – d) – (a – b – c – d)
= a + b – c – d – a + b + c + d (notice how the signs changed when we opened the second set of brackets)
= 2b (this is an even number).
If the difference between two numbers is even, can they have different parities?
No! So either both are even or both are odd.
Now, let us see what happens when a negative sign is switched to a positive sign.
Replace any negative sign in the expression a + b – c – d with a positive sign and find the difference between the two numbers.
What do you conclude from this observation?
Starting from any expression, we can get 7 expressions by switching one or more ‘+’ and ‘–’ signs. Thus, all the expressions have the same parity!

Explanation 2: We know that
odd ± odd = even
even ± even = even
odd ± even = odd.

We have seen that the parity of a + b and a – b is the same, regardless of the parities of a and b.
In short, a ± b have the same parity. By the same argument, a ± b + c and a ± b – c have the same parity.
Extending this further, we can say that all the expressions a ± b ± c ± d have the same parity.

Explanation 3: This can also be explained using the positive and negative token model you studied in the chapter on Integers. Try to think about how.

The number of ways to choose 4 numbers a, b, c, d, and combine them using ‘+’ and ‘–’ signs is infinite. Mathematical reasoning allows us to prove that all the combinations a ± b ± c ± d always have the same parity, without having to go through them one by one.

Several problems in mathematics can be thought about and solved in different ways. While the method you came up with may be dear to you, it can be amusing and enriching to know how others thought about it. Two tidbits: ‘share’ and ‘listen’.

Is the phenomenon of all the expressions having the same parity limited to taking 4 numbers? What do you think?

‘What if …?’, ‘Will it always happen?’ Wondering and posing questions, and conjectures are as much a part of mathematics as problem solving.

Breaking Even
We know how to identify even numbers. Without computing them, find out which of the following arithmetic expressions are even.

Using our understanding of how parity behaves under different operations, identify which of the following algebraic expressions give an even number for any integer values for the letter-numbers.

The expression 4m + 2q will always evaluate to an even number for any integer values of m and q. We can justify this in two different ways:
We know 4m is even and 2q is even for any integers m and q. Therefore, their sum will also be even.
The expression 4m + 2q is equal to the expression 2(2m + q).
Here, the expression 2(2m + q) means 2 times 2m + q.
In other words, 2 is a factor of this expression.
Therefore, this expression will always give an even number for any integers m and q.

For example, if m = 4 and q = -9, the expression 4m + 2q becomes 4 × 4 + 2 × (-9) = -2, which is an even number.
In the expression x² + 2, x² is even if x is even, and x2 is odd if x is odd.
Therefore, the expression x² + 2 will not always give an even number.
An example and a non-example of when the expression evaluates to an even number:

• if x = 6, then x² + 2 = 38
• if x = 3, then x² + 2 = 11.

Similarly, determine and explain which of the other expressions always give even numbers. Write a couple of examples and non-examples, as appropriate, for each expression. Write a few algebraic expressions that always give an even number.

Pairs to Make Fours
Take a pair of even numbers. Add them. Is the sum divisible by 4?
Try this with different pairs of even numbers.
When is the sum a multiple of 4, and when is it not?
Is there a general rule or a pattern?
Even numbers can be of two types based on the remainders they leave when divided by 4.

Even numbers that are multiples of 4 leave a remainder of 0 when divided by 4.

Even numbers that are not multiples of 4 leave a remainder of 2 when divided by 4

When will two even numbers add up to give a multiple of 4?
This problem is similar to the question of identifying when adding two numbers will result in an even number. Can you see this? There are three cases to examine:

What happens when we add a multiple of 4 to an even number that is not a multiple of 4?
Is it similar to the case of the parity of the sum of an even and an odd number?
Look at the following expressions and the visualisation. Write the corresponding explanation and examples.

Notice how we are able to generalise and prove properties of arithmetic using algebra and also using visualisation.

Always, Sometimes, or Never
We examine different statements about factors and multiples and determine whether a statement is ‘Always True’, ‘Sometimes True’, or ‘Never True’.
We know that the sum of any two multiples of 2 is also a multiple of 2.

1. If 8 exactly divides two numbers separately, it must exactly divide their sum.

Statement 1 is always true. Determine if it is true with subtraction.
In general, if a divides M and a divides N, then a divides M + N and a divides M – N.
In other words, if M and N are multiples of a, then M + N and M – N will also be multiples of a.

2. If a number is divisible by 8, then 8 also divides any two numbers (separately) that add up to the number.

So, statement 2 is sometimes true.

3. If a number is divisible by 7, then all multiples of that number will be divisible by 7.

The number 7jm or (7 × j × m) has a factor of 7.
We can see that Statement 3 is always true.

In general, if A is divisible by k, then all multiples of A are divisible by k.

4. If a number is divisible by 12, then the number is also divisible by all the factors of 12.

In general, if A is divisible by k, then A is divisible by all the factors of k.
Hence, Statement 4 is always true.

5. If a number is divisible by 7, then it is also divisible by any multiple of 7.

We can see that this statement is only sometimes true.

Examine each of the following statements, and determine whether it is ‛Always true’, ‛Sometimes true’, or ‛Never true’.

6. If a number is divisible by both 9 and 4, it must be divisible by 36.

7. If a number is divisible by both 6 and 4, it must be divisible by 24.

In general, if A is divisible by k and A is also divisible by m, then A is divisible by the LCM of k and m. This is because A is a multiple of k and also a multiple of m, so A’s prime factorisation should contain the prime factorisation of LCM (k, m).

8. When you add an odd number to an even number, we get a multiple of 6.

We know that multiples of 6 are all even numbers. The sum of an odd number and an even number will be an odd number. Therefore, this statement is never true. We can also explain this algebraically.
Suppose, (2n) + (2m + 1) = 6j,
where 2n is an even number, 2m + 1 is an odd number, and 6j is a multiple of 6.
Then 2n + 2m = 6j – 1
2(n + m) = 6j – 1, which means 2(n + m), which is an even number, should be equal to 6j – 1, which is an odd number. This is never true.

What Remains?
Find a number that has a remainder of 3 when divided by 5. Write more such numbers.
Which algebraic expression(s) capture all such numbers?
(i) 3k + 5
(ii) 3k – 5
(iii) \(\frac {3k}{5}\)
(iv) 5k + 3
(v) 5k – 2
(vi) 5k – 3
The numbers that leave a remainder of 0 when divided by 5 are the multiples of 5.
But we want numbers that leave a remainder of 3 when divided by 5.
These numbers are 3 more than multiples of 5.
Multiples of 5 are of the form 5k.
So, numbers that leave a remainder of 3 when divided by 5 are those of the form 5k + 3.

Let us consider another expression, 5k – 2, and see the values it takes for different values of k.

Numbers that leave a remainder of 3 when divided by 5 can also be seen as 2 less than multiples of 5; 5k – 2, where k ≥ 1.
Are there other expressions that generate numbers that are 3 more than a multiple of 5?

Checking Divisibility Quickly Class 8 Notes

Earlier, you have learnt shortcuts to check whether a given number, written in the Indian number system, is divisible by 2, 4, 5, 8, and 10. Let us revisit them.

Divisibility by 10, 5, and 2: If the units digit of a number is ‘0’, then it is divisible by 10. Let us understand why this works through algebra. We can write the general form of a number in the Indian system using a set of letter-numbers. For example, a 5-digit number can be expressed as edcba, denoting e × 10000 + d × 1000 + c × 100 + b × 10 + a. The letter-numbers e, d, c, b, and a denote each digit of a 5-digit number.

Any number can be written in general as … dcba, where the letter numbers a, b, c, and d represent the units, tens, hundreds, and thousands digit, respectively, and so on. As a sum of place values, this number is … +1000d + 100c + 10b + a.
(For example, in the number 4075, d = 4, c = 0, b = 7, and a = 5.)

We know that each place value, except the units place, is a multiple of 10. So, 10b, 100c, … all will be multiples of 10. Hence, the number will be divisible by 10 if and only if the units digit a is 0. Similarly, explain using algebra why the divisibility shortcuts for 5, 2, 4, and 8 work. Let us now examine shortcuts to check divisibility by some other numbers and explain why they work!

A Shortcut for Divisibility by 9
Can you say, without actually calculating, which of these numbers are divisible by 9: 999, 909, 900, 90, 990?
All of them.

Can we say that any number made up of only the digits ‘0’ and ‘9’, in any order, will always be divisible by 9?
Yes, if each digit is either 0 or 9, then each term in its expanded form will be 9 × __ or 0 × __ (the ‘__’ denotes a place value). This means each term will be a multiple of 9, for example,
99009 = 9 × 10000 + 9 × 1000 + 0 × 100 + 0 × 10 + 9 × 1.

But this shortcut alone cannot identify all the multiples of 9. Unlike the numbers 2, 5, and 10, we cannot identify the multiples of 9 by just looking at the unit’s digit. 99 and 109 are two numbers with 9 as the units digit, but 99 is divisible by 9, while 109 is not.

Is 10 divisible by 9? If not, what is the remainder?
Check the divisibility of other multiples of 10 (10, 20, 30, …) by 9.
You will notice that for any multiple of 10, the remainder is the same as the number of tens.
Similarly, look at the remainder when the multiples of 100 (100, 200, 300, … ) are divided by 9. What do you notice?
The remainder is the same as the number of hundreds for any multiple of 100.
Using this observation, find the remainder when 427 is divided by 9.

We see that 427 has 4 hundreds; thus, its corresponding remainder (upon division by 9) would be 4. 427 has 2 tens, and its corresponding remainder would be 2. We have 7 units remaining. Adding all the remainders, we get 4 + 2 + 7 = 13. We can make one more group of 9 with 13, leaving a remainder of 4. Therefore, 427 ÷ 9 gives a remainder of 4.

Will this work with bigger numbers?
You can see that this is true for any place value:
1 = 0 + 1
10 = 9 + 1
100 = 99 + 1
1000 = 999 + 1
10000 = 9999 + 1, and so on.

Each digit thus denotes the remainder when the corresponding place value is divided by 9.
For example, to find the remainder of 7309 when divided by 9, we can just add all the digits 7 + 3 + 0 + 9 to get 19. This can be seen as follows:

This means that the number 7309 is 19 more than some multiple of 9. The digits 1 and 9 can further be added to get 1 + 9 = 10.
Now, we can say that 7309 is 10 more than a multiple of 9.
And repeating this step for the number 10, we get the remainder to be 1 + 0 = 1, meaning 7309 is 1 more than a multiple of 9.
Therefore, 7309 ÷ 9 gives a remainder of 1.
A number is divisible by 9 if and only if the sum of its digits is divisible by 9.
Also, we can add the digits of a number repeatedly till a single digit is obtained.
This single digit is the remainder when the number is divided by 9.

Look at each of the following statements. Which are correct and why?

• If a number is divisible by 9, then the sum of its digits is divisible by 9.
• If the sum of the digits of a number is divisible by 9, then the number is divisible by 9.
• If a number is not divisible by 9, then the sum of its digits is not divisible by 9.
• If the sum of the digits of a number is not divisible by 9, then the number is not divisible by 9.

A Shortcut for Divisibility by 3
We know that all the multiples of 9 are also multiples of 3. That is, if a number is divisible by 9, it will also be divisible by 3.
However, other multiples of 3 are not multiples of 9, for example, 5, 33, and 87.

The shortcut to find the divisibility by 3 is similar to the method for 9.
A number is divisible by 3 if the sum of its digits is divisible by 3.
Explore the remainders when powers of 10 are divided by 3. Explain why this method works.

A Shortcut for Divisibility by 11
Interestingly, the shortcut for 11 is also based on checking the remainders with place value. Let us see how.

This alternating pattern of one more than 11 and one less than 11 continues for higher place values.
Since 400 contains 4 hundreds, 400 is 4 more than a multiple of 11 (396 + 4).
Since 60 contains 6 tens, 60 is 6 less than a multiple of 11 (66 – 6).
Since 2 contains 2 units, 2 is 2 more than a multiple of 11, i.e., 2 = (0 + 2).

Using these observations, can you tell whether the number 462 is divisible by 11?
What could be a general method or shortcut to check divisibility by 11?

We saw that the place values alternate as 1 more and 1 less than a multiple of 11. Using this observation,

The difference between these two sums 8 – 11 = -3, indicating that the number 3,28,105 is 3 short of or 8 more than a multiple of 11.

Look at the following procedure.

Is this method similar to or different from the method we saw just before?
Fill in the following table. Find a quick way to do this?

More on Divisibility Shortcuts

Divisibility Shortcuts for Other Numbers
How can we find out if a number is divisible by 6?
Will checking its divisibility by its factors 2 and 3 work?
Use the shortcuts for 2 and 3 on these numbers and divide each number by 6 to verify 38, 225, 186, and 64.
How about checking divisibility by 24? Will checking the divisibility by its factors, 4 and 6, work? Why or why not?

Determining divisibility by 24 by checking divisibility by 4 and by 6 does not work.
For example, the number 12 is divisible by both 4 and 6, but not by 24.
To check for the divisibility by 24, we can instead check for the divisibility by 3 and the divisibility by 8.

Explain using prime factorisation why checking divisibility by 3 and 8 works for checking divisibility by 24, but checking divisibility by 4 and 6 is not sufficient for checking divisibility by 24. There are such shortcuts to check divisibility by every number until 100, and for some numbers beyond 100. You may try to understand how these work after learning certain concepts in higher grades.

Digital Roots
Take a number. Add its digits repeatedly till you get a single-digit number. This single-digit number is called the digital root of the number.
For example, the digital root of the number 489710 will be 2(4 + 8 + 9 + 7 + 1 + 0 = 29, 2 + 9 = 11, 1 + 1 = 2).

What property do you think this digital root will have? Recall that we did this while finding the divisibility shortcut for 9.
Between the numbers 600 and 700, which numbers have the digital root: (i) 5, (ii) 7, (iii) 3?
Write the digital roots of any 12 consecutive numbers. What do you observe?
We saw that the digital root of multiples of 9 is always 9.
Now, find the digital roots of some consecutive multiples of (i) 3, (ii) 4, and (iii) 6.
What are the digital roots of numbers that are 1 more than a multiple of 6? What do you notice?
Try to explain the patterns noticed.
I’m made of digits, each tiniest and odd,
No shared ground with root #1 – how odd!
My digits count, their sum, my root – All point to one bold number’s pursuit – The largest odd single-digit I proudly claim. What’s my number? What’s my name?

Aryabhata II’s (c. 950 CE) work, Mahasiddhanta, mentions the method of computing the digital root of a number by repeatedly adding the digits till a single-digit number is obtained. This method is known to have been used to perform checks on calculations of arithmetic operations.

Digits in Disguise Class 8 Notes

Last year, we saw cryptarithm puzzles where each letter stands for a digit, each digit is represented by at most one letter, and the first digit of a number is never 0.

Solve the cryptarithms given below.

Let us now try solving some cryptarithms involving multiplication.
(v) PQ × 8 = RS.
Guna says, “Oh, this means a 2-digit number multiplied by 8 should give another 2-digit number.
I know that 10 × 8 = 80. But the units digits of 10 and 80 are the same, which we don’t want.
For the same reason, PQ cannot be 11 as P and Q correspond to different digits. 12 × 8 = 96 fits all the conditions”. Can PQ be 13? Think.
It is not possible because 13 × 8 = 104. For all 2-digit numbers greater than 12, the product with 8 is a 3-digit number.

(vi) Try this now: GH × H = 9K.
This means a 2-digit number multiplied by a 1-digit number gives another 2-digit number in the 90s. Observe the letters corresponding to the unit digits in this cryptarithm. Pick the solution to this question from the options given below:
11 × 9 = 99, 12 × 8 = 96, 46 × 2 = 92, 24 × 4 = 96, 47 × 2 = 94, 31 × 3 = 93, 16 × 6 = 96.

(vii) Here is one more: BYE × 6 = RAY.
Anshu says, “Since the product is a 3-digit number, B can’t be 2 or more. If B = 2, i.e., 2 hundreds, the product will be more than 1200. So, B = 1.”

What can you say about ‘Y’? What digits are possible/not possible?
“Y cannot be 7 or more because, if Y = 7, then 170 × 6 = 1020; but we want a 3-digit product. Also, Y will be even”, Anshu explains. We can solve cryptarithms using patterns, properties, and reasoning related to numbers and operations.

The post Number Play Class 8 Notes Maths Chapter 5 appeared first on Learn CBSE.

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