Number Play Class 8 Solutions Ganita Prakash Maths Chapter 5

Students often refer to Ganita Prakash Class 8 Solutions and Class 8 Maths Chapter 5 Number Play NCERT Solutions Question Answer to verify their answers.

NCERT Class 8 Maths Chapter 5 Number Play Solutions Question Answer

Ganita Prakash Class 8 Chapter 5 Solutions Number Play

NCERT Class 8 Maths Ganita Prakash Chapter 5 Number Play Solutions Question Answer

5.1 Is This a Multiple Of?

Sum of Consecutive Numbers

NCERT In-Text Questions (Pages 112-115)

Question 1.
Explore these questions and any others that may occur to you. Discuss them with the class.
Solution:
Do it yourself.

Question 2.
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.

Solution:

Question 3.
Evaluate each expression and write the result next to it. Do you notice anything interesting?
Solution:
1. 3 + 4 + 5 + 6 = 18
2. 3 + 4 + 5 – 6 = 6
3. 3 + 4 – 5 + 6 = 8
4. 3 + 4 – 5 – 6 = -4
5. 3 – 4 + 5 + 6 = 10
6. 3 – 4 + 5 – 6 = -2
7. 3 – 4 – 5 + 6 = 0
8. 3 – 4 – 5 – 6 = -12
All results are even numbers.

Question 4.
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?
Solution:
Do it yourself.

Question 5.
Repeat this for one more set of 4 consecutive numbers. Share your findings.

Solution:


Again, all results are even.

Question 6.
Replace any negative sign in the expression a + b – c – d with a positive sign and find the difference between the two numbers.
Solution:
Consider one of the expressions a + b – c – d.
Replacing -c by +c, we get a + b + c – d.
It has changed by (a + b – c – d) – (a + b + c – d) = a + b – c – d – a – b – c + d = -2c (This is an even number)
So, if the difference between two numbers is even, then they both are even or both are odd.

Question 7.
Is the phenomenon of all the expressions having the same parity limited to taking 4 numbers? What do you think?
Solution:
Do it yourself.

Breaking Even

NCERT In-Text Questions (Pages 115-116)

Question 1.
We know how to identify even numbers. Without computing them, find out which of the following arithmetic expressions are even.

Solution:
43 + 37 = Even [Odd + Odd = Even]
672 – 348 = Even [Even – Even = Even]
4 × 347 × 3 = Even [Even × Odd × Odd = Even]
708 – 477 = Odd [Even – Odd = Odd]
809 + 214 = Odd [Odd + Even = Odd]
119 × 303 = Odd [Odd × Odd = Odd]
543 – 479 = Even [Odd – Odd = Even]
5133 = Odd [(Odd)Odd = Odd]

Question 2.
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.

Solution:
2a + 2b = Even [2 × Even/Odd + 2 × Even/Odd = Even + Even = Even]
3g + 5h = Even/Odd [3 × Even/Odd + 5 × Even/Odd = Even/Odd + Even/Odd = Even/Odd]
4m + 2n = Even [4 × Even/Odd + 2 × Even/Odd = Even + Even = Even]
2u – 4v = Even [2 × Even/Odd – 4 × Even/Odd = Even – Even = Even]
13k – 5k = Even [(13 – 5) × Even/Odd = Even × (Even/Odd) = Even]
6m – 3n = 3 × [2m – n] = 3 × [Even – Even/Odd] = 3 × [Even/Odd] = Even/Odd
x² + 2 = (Even/Odd)² + 2 = Even/Odd + 2 = Even/Odd
b² + 1 = (Even/Odd)² + 1 = Even/Odd + 1 = Odd/Even
4k × 3j = Even × (Even/Odd) = Even
Therefore, the algebraic expressions 2a + 2b, 4m + 2n, 2u – 4v, 13k – 5k, and 4k × 3j give an even number for any integer value for the letter-numbers.

Question 3.
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.
Solution:
Consider 2u – 4v.
Here, 2u is even, 4v is even for any integers u and v.
So, their difference will also be even.
The expression 2u – 4v is equal to the expression 2(u – 2v), which means 2 times u – 2v.
As 2 is a factor of this expression, this expression will always give an even number for any integers u and v.
In the expression b² + 3, b² is even if b is even, and b² is odd if b is odd.
So, b² + 3 will not always give an even number.
An example and a non-example of when the expression evaluates to an even number are as follows:

• If b = 6, then b² + 3 = 39
• If b = 7, then b² + 3 = 52.

Question 4.
Write a few algebraic expressions that always give an even number.
Solution:
A few algebraic expressions that always give an even number are: 2x + 2y, 6x – 4m, 8p × q.

Pairs to Make Fours

NCERT In-Text Questions (Pages 116-117)

Question 1.
When will two even numbers add up to give a multiple of 4?
Solution:
Two even numbers will add up to a multiple of 4 if and only if both are either Multiples of 4 or even but not multiples of 4.

Question 2.
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?
Solution:
When we add a multiple of 4 to an even number that is not a multiple of 4, the result is still an even number, but not a multiple of 4.
Like: 4y(multiple of 4) + (4x + 2) = 4y + 4x + 2, is not a multiple of 4.
Yes, it is similar to the case of parity of the sum of an even and an odd number.

Question 3.
Look at the following expressions and the visualisation. Write the corresponding explanation and examples.

Solution:

Always, Sometimes, or Never

NCERT In-Text Questions (Pages 118-121)

Question 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.
Solution:
Two numbers with multiples of 8 are 8a and 8b.
Now, their difference = 8a – 8b = 8(a-b), which is also a multiple of 8.
Thus, if 8 exactly divides two numbers, it always exactly divides both their sum and their difference.
Examples:

• 16 – 8 = 8
• 120 – 80 = 40
• 56 – 16 = 40

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

Question 6.
If a number is divisible by both 9 and 4, it must be divisible by 36.
Solution:
LCM(9, 4) = 36.
Now, consider the number 72, which is divisible by 9 and 4, and it is also divisible by LCM (9, 4), i.e., 36.
∴ If a number is divisible by 9 and 4, then it must be divisible by 36.
So, the statement is ‘always true’.

Question 7.
If a number is divisible by both 6 and 4, it must be divisible by 24.
Solution:
LCM(6, 4) = 12
Consider the number 36 multiple of 12, which is divisible by both 6 and 4 but not by 24.
Consider another number, 48, which is divisible by both 6 and 4, and it is also divisible by 24.
So, the statement is ‘sometimes true’.

What Remains?

NCERT In-Text Questions (Pages 121-122)

Question 1.
Find a number that has a remainder of 3 when divided by 5. Write more such numbers.
Solution:
Such numbers can be written as: 5k + 3, where k is any whole number.
Examples: 3, 8, 13, 18, 23,….

Question 2.
Are there other expressions that generate numbers that are 3 more than a multiple of 5?
Solution:
Yes, there are infinitely many such expressions that are 3 more than a multiple of 5.
For example: 10x + 3, 5(2x + 1) – 2, 25x + 3, where x is any whole number.

Figure it Out (Pages 122-123)

Question 1.
The sum of four consecutive numbers is 34. What are these numbers?
Solution:
Let the numbers be x, x + 1, x + 2, x + 3.
Sum = x + (x + 1) + (x + 2) + (x + 3) = 4x + 6 ……(i)
Given, the sum of four consecutive numbers = 34
⇒ 4x + 6 = 34 [Using (i)]
⇒ 4x = 28
⇒ x = 7
So, the required numbers are 7, 8, 9, and 10.

Question 2.
Suppose p is the greatest of five consecutive numbers. Describe the other four numbers in terms of p.
Solution:
Given, ‘p’ is the greatest of five consecutive numbers.
So, the other four numbers in terms of p are p – 4, p – 3, p – 2, and p – 1.

Question 3.
For each statement below, determine whether it is always true, sometimes true, or never true. Explain your answer. Mention examples and non-examples as appropriate. Justify your claim using algebra.
(i) The sum of two even numbers is a multiple of 3.
(ii) If a number is not divisible by 18, then it is also not divisible by 9.
(iii) If two numbers are not divisible by 6, then their sum is not divisible by 6.
(iv) The sum of a multiple of 6 and a multiple of 9 is a multiple of 3.
(v) The sum of a multiple of 6 and a multiple of 3 is a multiple of 9.
Solution:
(i) Consider 2 + 4 = 6, which is a multiple of 3.
2 + 8 = 10, which is not a multiple of 3.
So, the given statement is sometimes true.
Algebraic justification: Let the two even numbers be 2a and 2b.
Sum = 2(a + b), which is even but not necessarily divisible by 3.

(ii) Let us consider a number: 12, which is not divisible by 18. It is also not divisible by 9.
Let us consider another number: 9 is not divisible by 18, but is divisible by 9.
Algebraic justification: Let the number be 21k, which is not divisible by 18, but is divisible by 9, for k = 1, 3, 5,…
So, the given statement is sometimes true.

(iii) 4 and 8 are not both divisible by 6, but their sum, i.e., 4 + 8 = 12, is divisible by 6.
Algebraic justification: The number, which is not divisible by 6, is in the form 6k + r, where r = 1, 2, 3, 4, or 5.
And sum of two numbers = (6k₁ + r₁) + (6k₂ + r₂) = 6(k₁ + k₂) + (r₁ + r₂) is divisible by 6, when r₁ + r₂ = 6.
So, the given statement is sometimes true.

(iv) Let the multiple of 6 be 6m and the multiple of 9 be 9n.
Sum = 6m + 9n = 3(2m + 3n), which is a multiple of 3.
So, the given statement is always true.

(v) Consider, 6 + 3 = 9, i.e., the sum of a multiple of 6 and a multiple of 3 is a multiple of 9.
Consider 6 + 6 = 12, which is not a multiple of 9.
Algebraic justification: The sum of multiples of 6 and multiples of 3 is the form 6i + 3j = 3(2i + j), which multiple of 3,
But when i = j, or (2i + j) is a multiple of 3, then it is a multiple of 9.
So, the given statement is sometimes true.

Question 4.
Find a few numbers that leave a remainder of 2 when divided by 3 and a remainder of 2 when divided by 4. Write an algebraic expression to describe all such numbers.
Solution:
Consider the numbers 2, 14, 26, 38,… that leave a remainder of 2 when divided by 3 and a remainder of 2 when divided by 4.
So, the algebraic expression is 12k + 2, where k is any whole number.

Question 5.
“I hold some pebbles, not too many, when I group them in 3’s, one stays with me. Try pairing them up — it simply won’t do. A stubborn odd pebble remains in my view. Group them by 5, yet one’s still around, but grouping by seven, perfection is found. More than one hundred would be far too bold. Can you tell me the number of pebbles I hold?”

Solution:
Clues are:
Leaves 1 when pebbles are divided by 3

• Cannot be divided into equal parts, so they are odd.
• Leaves 1 pebble when divided by 5.
• Leaves 0 pebble when divided by 7.
• Pebbles are less than 100.

Multiples of 7 under 100 are 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98.
So, 91 is the number of pebbles that is divisible by 7, leaves a remainder of 1 when divided by 3 and by 5, and it is less than 100.

Question 6.
Tathagat has written several numbers that leave a remainder of 2 when divided by 6. He claims, “If you add any three such numbers, the sum will always be a multiple of 6.” Is Tathagat’s claim true?
Solution:
The number that Tathagat wrote can be in the form of 6k + 2, where k is an integer.
i.e., 6(0) + 2 = 2
6(1) + 2 = 8
6(2) + 2 = 14
6(3) + 2 = 20, and so on.
Now, if we add any of the above 3 numbers, that will be of the form (6k₁ + 2) + (6k₂ + 2) + (6k₃+ 2)
= 6(k₁ + k₂ + k₃) + 6
= 6(k₁ + k₂ + k₃ + 1), which is a multiple of 6.
Thus, Tathagat’s claim is true.

Question 7.
When divided by 7, the number 661 leaves a remainder of 3, and 4779 leaves a remainder of 5. Without calculating, can you say what remainders the following expressions will leave when divided by 7? Show the solution both algebraically and visually.
(i) 4779 + 661
(ii) 4779 – 661
Solution:
(i) The expression 4779 + 661, when divided by 7, leaves a remainder of 1.
[∵ Remainder of 4779 + remainder of 661 = 5 + 3 = 8 and when 8 ÷ 7 = 1 (remainder)]
Algebraically: Let 661 be represented as 7a + 3 and 4779 as 7b + 5, then we have
4779 + 661 = (7b + 5) + (7a + 3)
= 7(a + b) + 8
= 7(a + b) + 7 + 1
= 7(a + b + 1) + 1
Now, 7(a + b + 1) + 1 will leave a remainder of 1 when divided by 7.
Thus, 4779 + 661 will leave a remainder of 1.
For visually: Do it yourself.

(ii) The expression 4779 – 661, when divided by 7, leaves a remainder of 2.
[∵ Remainder of 4779 – remainder of 661 = 5 – 3 = 2 and when 2 ÷ 7 = 2(remainder)]
Algebraically: 4779 – 661 = (7b + 5) – (7a + 3)
= 7b – 7a + 5 – 3
= 7(b – a) + 2
Now, 7(b – a) + 2 will leave a remainder of 2 when divided by 7, and hence 4779 – 661 will leave a remainder of 2 when divided by 7.
For visually: Do it yourself.

Question 8.
Find a number that leaves a remainder of 2 when divided by 3, a remainder of 3 when divided by 4, and a remainder of 4 when divided by 5. What is the smallest such number? Can you give a simple explanation of why it is the smallest?
Solution:
Given the number leaves a remainder 2 when divided by 3, a remainder of 3 when divided by 4, and a remainder of 4 when divided by 5.
Given, in each case remainder is 1 less than the divisor.
So, N + 1 will be perfectly divisible by 3, 4, and 5.
LCM (3, 4, 5) = 60.
So, the smallest common multiple of 3, 4, and 5 is 60.
∴ N + 1 = 60
⇒ N = 59
So, the smallest number is 59, which leaves a remainder of 2 when divided by 3, a remainder of 3 when divided by 4, and a remainder of 4 when divided by 5.
Also, 59 is the smallest, because N + 1 is a multiple of 3, 4, and 5, and the smallest such multiple of 3, 4, and 5 is 60.

5.2 Checking Divisibility Quickly

NCERT In-Text Questions (Page 123)

Question 1.
Similarly, explain using algebra why the divisibility shortcuts for 5, 2, 4, and 8 work.
Solution:
Consider a number dcba written in the Indian numeral system can be expressed as 1000d + 100c + 10b + a.
Divisibility Shortcut for 5:
Here, 1000d, 100c, 10b are divisible by 5.
Hence, the number will be divisible by 5 if the units digit a is either ‘0’ or ‘5’.

Divisibility Shortcut for 2:
In the above numbers 1000d, 100c, 10b are divisible by 2.
Hence, the number will be divisible by 2 if a is even.

Divisibility Shortcut for 4:
In the above number 1000d, 100c are divisible by 4.
Hence, the number will be divisible by 4 if ‘10b + a’
i.e, the number formed by the last two digits is divisible by 4.

Divisibility Shortcut for 8:
In the above number 1000d is divisible by 8.
Now, the number will be divisible by 8 if ‘100c + 10b + a’
i.e, the number formed by the last three digits is divisible by 8.

A Shortcut for Divisibility by 9

NCERT In-Text Questions (Pages 123-126)

Question 1.
Is 10 divisible by 9? If not, what is the remainder?
Solution:
No, 10 is not divisible by 9.
On dividing 10 by 9, we will get a remainder of 1.
On dividing 20 by 9, we will get a remainder of 2.
On dividing 30 by 9, we will get a remainder of 3.
On dividing 40 by 9, we will get a remainder of 4.
So, for multiples of 10, the remainder when divided by 9 is the same as the number of tens,
i.e., remainder = number of tens digit.

Question 2.
Look at each of the following statements. Which are correct and why?
(i) If a number is divisible by 9, then the sum of its digits is divisible by 9.
(ii) If the sum of the digits of a number is divisible by 9, then the number is divisible by 9.
(iii) If a number is not divisible by 9, then the sum of its digits is not divisible by 9.
(iv) If the sum of the digits of a number is not divisible by 9, then the number is not divisible by 9.
Solution:
(i) This statement is correct, since a number is divisible by 9 if and only if the sum of its digits is divisible by 9.
(ii) This statement is correct, since a number is divisible by 9 if and only if the sum of its digits is divisible by 9.
(iii) This statement is correct because a number is divisible by 9 only when the sum of its digits is divisible by 9.
If the sum of digits is not divisible by 9, then the number is not divisible by 9.
(iv) Do it yourself.

Figure it Out (Page 126)

Question 1.
Find, without dividing, whether the following numbers are divisible by 9.
(i) 123
(ii) 405
(iii) 8888
(iv) 93547
(v) 358095
Solution:
(i) 123
Sum of digits = 1 + 2 + 3 = 6, which is not divisible by 9.
So, 123 is not divisible by 9.

(ii) 405
Sum of digits = 4 + 0 + 5 = 9, which is divisible by 9.
So, 405 is divisible by 9.

(iii) 8888
Sum of digits = 8 + 8 + 8 + 8 = 32, which is not divisible by 9.
So, 8888 is not divisible by 9.

(iv) 93547
Sum of digits = 9 + 3 + 5 + 4 + 7 = 28, which is not divisible by 9.
So, 93547 is not divisible by 9.

(v) 358095
Sum of digits = 3 + 5 + 8 + 0 + 9 + 5 = 30, which is not divisible by 9.
So, 358095 is not divisible by 9.

Question 2.
Find the smallest multiple of 9 with no odd digits.
Solution:
The smallest multiple of 9 with no odd digits means it uses only even digits, i.e., 0, 2, 4, 6, 8,…, is 288.

Question 3.
Find the multiple of 9 that is closest to the number 6000.
Solution:
6000 = 6 × 1000
= 6 × (999 + 1)
= 6 × 999 + 6 × 1
= 5994 (Multiple of 9) + 6 (Remainder)
Since 5994 is a multiple of 9 and 5994 + 9 = 6003 is the next multiple of 9 after 5994.
Therefore, the multiple of 9 closest to 6000 is 6003.

Question 4.
How many multiples of 9 are there between the numbers 4300 and 4400?
Solution:
We have two numbers, 4300 and 4400.

So, the multiples of 9 between 4300 and 4400 are 4302, 4311, 4320, 4329, 4338, 4347, 4356, 4365, 4374, 4383, 4392, i.e., 11.

A Shortcut for Divisibility by 3

NCERT In-Text Questions (Page 126)

Question 1.
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.
Solution:
We are considering the power of 10.
100 = 1. The remainder is 1 when divided by 3.
101 = 3 × 3 + 1. The remainder is 1 when divided by 3.
102 = 100 = 3 × 33 + 1. The remainder is 1 when divided by 3.
Thus, all powers of 10 have a remainder of 1 when divided by 3.
Now, let bcde be a number, we can write it using place value as 1000b + 100c + 10d + 1e
We can write the power of 10 as a multiple of 3 plus 1.
So, we have 1000b + 100c + 10d + 1e = (3k₁ + 1)b + (3k₂ + 1)c + (3k₃ + 1)d + (3k₄ + 1)e
for some integers k₁, k₂, k₃, k₄.
⇒ bcde = 3(k₁b + k₂c + k₃d + k₄e) + (b + c + d + e)
⇒ bcde is divisible by 3 if (b + c + d + e) is divisible by 3.
Thus, a number is divisible by 3 if the sum of its digits is divisible by 3.

A Shortcut for Divisibility by 11

NCERT In-Text Questions (Pages 126-129)

Question 1.
Using these observations, can you tell whether the number 462 is divisible by 11?
Solution:
First, we will break down 462, i.e., 462 = 400 + 60 + 2
Now, 400 = 396 + 4 = 11 × 36 + 4
400 is 4 more than a multiple of 11.
60 = 66 – 6 = 6 × 11 – 6
60 is 6 less than a multiple of 11.
2 = 0 + 2 = 11 × 0 + 2
2 is 2 more than a multiple of 11.
Now, we apply alternating sign to combine these differences: 4 – 6 + 2 = 0
∴ The total is 0, a multiple of 11. So, 462 is divisible by 11.

Question 2.
If this difference is 11 or a multiple of 11, what does that say about the remainder obtained when the number is divisible by 11?
Solution:
If the difference is 11 or a multiple of 11, then the remainder is 0, and the number is divisible by 11.

Question 3.
Using this shortcut, find out whether the following numbers are divisible by 11. Further, find the remainder if the number is not divisible by 11.
(i) 158
(ii) 841
(iii) 481
(iv) 5529
(v) 90904
(vi) 857076
Look at the following procedure.

Solution:
(i) 158
Total excess = 1 + 8 = 9
Total short = 5
Difference = 9 – 5 = 4, not divisible by 11.
∴ Remainder = 4.

(ii) 841
Total excess = 8 + 1 = 9
Total short = 4
Difference = 9 – 4 = 5, not divisible by 11.
∴ Remainder = 5.

(iii) 481
Total excess = 4 + 1 = 5
Total short = 8
Difference = 5 – 8 = -3
-3 indicating that the number 481 is 3 short or 8 more than a multiple of 11.
So, the remainder is 8. (∵ Here, a multiple of 11 cannot be more than 481)

(iv) 5529
Total excess = 5 + 9 = 14
Total short = 5 + 2 = 7
Difference = 14 – 7 = 7, not divisible by 11.
∴ Remainder = 7.

(v) 90904
Total excess = 9 + 9 + 4 = 22
Total short = 0 + 0 = 0
Difference = 22 – 0 = 22, divisible by 11.
∴ Remainder = 0.

(vi) 857076
Total excess = 5 + 0 + 6 = 11
Total short = 8 + 7 + 7 = 22
Difference = 11 – 22 = -11, divisible by 11.
∴ Remainder = 0.

Question 4.
Is this method similar to or different from the method we saw just before?
Solution:
The first method (place value groups) is based on place value, i.e., 1, 10, 100, etc.
Then the operation in groups, and then subtract the two sums.
In the second method. (Alternate signs) It is based on digit position starting from the unit.
Then the operation is alternating addition and subtraction.
It always alternates from the right.
Thus, we can say that both methods are mathematically equivalent but methodologically different.

Question 5.
Fill in the following table. Find a quick way to do this?

Solution:

More on Divisibility Shortcuts

Divisibility Shortcuts for Other Numbers

NCERT In-Text Questions (Pages 129-130)

Question 1.
How can we find out if a number is divisible by 6?
Solution:
Prime factors of 6 are 2 and 3.
So, if a number is divisible by both 2 and 3, then it must be divisible by 6.

Question 2.
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, 64.
Solution:

Question 3.
How about checking divisibility by 24? Will checking the divisibility by its factors, 4 and 6, work? Why or why not?
Solution:
Prime factorisation of 24 = 2 × 2 × 2 × 3 = 2³ × 3
Now, 4 = 2² and 6 = 2 × 3
So, if a number is divisible by both 4 and 6, it is divisible by 2² and (2 × 3).
So it covered just 2² × 3 = 12.
But divisible by 3 and 8 covers 3 × 2³ = 24.
So, divisibility by 3 and 8 works for checking divisibility by 24.

Digital Roots

NCERT In-Text Questions (Page 130)

Question 1.
What property do you think this digital root will have? Recall that we did this while finding the divisibility shortcut for 9.
Solution:
The digital root is the remainder when a number is divided by 9, except that when the remainder is 0, the digital root is 9.

Question 2.
Between the numbers 600 and 700, which numbers have the digital root: (i) 5, (ii) 7, (iii) 3?
Solution:
Let us consider the number:
601 → 6 + 0 + 1 = 7 → Digital Root 7
602 → 6 + 0 + 2 = 8
603 → 6 + 0 + 3 = 9
604 → 6 + 0 + 4 = 10 = 1 + 0 = 1
605 → 6 + 0 + 5 = 11 = 1 + 1 = 2
606 → 6 + 0 + 6 = 12 = 1 + 2 = 3 → Digital Root 3
607 → 6 + 0 + 7 = 13 = 1 + 3 = 4
608 → 6 + 0 + 8 = 14 = 1 + 4 = 5 → Digital Root 5
609 → 6 + 0 + 9 = 15 = 1 + 5 = 6
610 → 6 + 1 + 0 = 7 → Digital Root 7
Here, we can see that the digital roots are repeating themselves after each 9th number. Therefore,

Question 3.
Write the digital roots of any 12 consecutive numbers. What do you observe?
Solution:
Let us consider the numbers 101 – 112.
101 → 1 + 0 + 1 = 2
102 → 1 + 0 + 2 = 3
103 → 1 + 0 + 3 = 4
104 → 1 + 0 + 4 = 5
105 → 1 + 0 + 5 = 6
106 → 1 + 0 + 6 = 7
107 → 1 + 0 + 7 = 8
108 → 1 + 0 + 8 = 9
109 → 1 + 0 + 9 = 10 = 1 + 0 = 1
110 → 1 + 1 + 0 = 2
111 → 1 + 1 + 1 = 3
112 → 1 + 1 + 2 = 4
We observe that digital roots repeat in a cycle of 9, because there are only possible non-zero digital roots (1 through 9).
After 9 steps, the pattern repeats from step 1 onwards.

Question 4.
Now, find the digital roots of some consecutive multiples of (i) 3, (ii) 4, and (iii) 6.
Solution:
(i) Consecutive multiples of 3 are 3, 6, 9, 12, 15, 18, 21,…..
3 → Digital Root 3
6 → Digital Root 6
9 → Digital Root 9
12 → Digital Root 3
15 → Digital Root 6
18 → Digital Root 9
21 → Digital Root 3

(ii) Consecutive multiples of 4 are 4, 8, 12, 16, 20, 24,…
4 → Digital Root 4
8 → Digital Root 8
12 → Digital Root 3
16 → Digital Root 7
20 → Digital Root 2
24 → Digital Root 3

(iii) Consecutive multiples of 6 are 6, 12, 18, 24, 30, 36,…..
6 → Digital Root 6
12 → Digital Root 3
18 → Digital Root 9
24 → Digital Root 6
30 → Digital Root 3
36 → Digital Root 9

Question 5.
What are the digital roots of numbers that are 1 more than a multiple of 6? What do you notice?
Solution:
Numbers that are 1 more than a multiple of 6 are 7, 13, 19, 25, 31, 37, and so on.

The digital roots repeat in a cycle of three: 7, 4, 1,…

Question 6.
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?

Solution:
All digits odd, i.e., 1, 3, 5, 7, 9,…
The sum of digits should be a number whose digital root is the largest odd single digit, i.e., 9.
Also, root #1 is excluded
So, the possible number is 111111111
Here, 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 9
So, the number is 111111111 and the name is “Eleven crore eleven lakh eleven thousand one hundred eleven”.

Figure it Out (Page 131)

Question 1.
The digital root of an 8-digit number is 5. What will be the digital root of 10 more than that number?
Solution:
Let the 8-digit number be N.
Whose digital root is 5.
Now, 10 more than the number = N + 10
So, the digital root of 10 more than that number = 5 + 1 = 6.

Question 2.
Write any number. Generate a sequence of numbers by repeatedly adding 11. What would be the digital roots of this sequence of numbers? Share your observations.
Solution:
Consider the numbers like 100, 111, 122, 133, 144,…

Thus, digital roots of this sequence follow a cycle 1, 3, 5, 7, 9, 2, 4, 6, 8,… i.e., repeating cycle of length 9.
Starting with 1, the digital root increases by 2 at each step.

Question 3.
What will be the digital root of the number 9a + 36b + 13?
Solution:
Consider 9a + 36b + 13
9a → digital root is 9 (since 9 × any number is a multiple of 9).
36b → digital root is 9
13 → 1 + 3 = 4 → digital root is 4.
So digital root of 9a + 36b + 13 = 9 + 9 + 4 = 22 → 2 + 2 = 4.

Question 4.
Make conjectures by examining if there are any patterns or relations between
(i) the parity of a number and its digital root.
(ii) the digital root of a number and the remainder obtained when the number is divided by 3 or 9.
Solution:
(i) Even numbers can have odd or even digital roots.
(E.g. 20 → 2, 12 → 3, 14 → 5)
Odd numbers can also have odd or even digital roots.
(E.g. 13 → 4, 133 → 7)
So, parity of digital roots does not match with parity of numbers.

(ii) The digital root of a number is equal to the remainder when the number is divided by 9, unless the remainder is 0, then the digital root is 9.
If the digital root = 3, 6, then the number is divisible by 3.
Thus, digital root helps us in identifying the remainder when the number is divided by 9 or 3 (or gives information about divisibility by 9 or 3).

5.3 Digits in Disguise

NCERT In-Text Questions (Pages 131-132)

Question 1.
Solve the cryptarithms given below.

Solution:

Question 2.
(vi) Try this now: GH × H = 9K. 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.
Solution:
The solution is 24 × 4 = 96

Question 3.
(vii) What can you say about ‘Y’? What digits are possible/not possible?
Solution:
As

Therefore, B = 1, Y = 0, E = 5, R = 6, A = 3

Question 4.
Solve the following:
(i) UT × 3 = PUT
(ii) AB × 5 = BC
(iii) L2N × 2 = 2NP
(iv) XY × 4 = ZX
(v) PP × QQ = PRP
(vi) JK × 6 = KKK
Solution:

Figure it Out (Pages 132-134)

Question 1.
If 31z5 is a multiple of 9, where z is a digit, what is the value of z? Explain why there are two answers to this problem.
Solution:
Given 31z5 is a multiple of 9.
So, 3 + 1 + z + 5 = 9 + z is divisible by 9.
So, z = 0 or 9.
∴ The required numbers are 3105 or 3195.
Now, there are two answers to this problem because both 3105 and 3195 have digital roots 9, which is divisible by 9.

Question 2.
“I take a number that leaves a remainder of 8 when divided by 12. I take another number, which is 4 short of a multiple of 12. Their sum will always be a multiple of 8”, claims Snehal. Examine his claim and justify your conclusion.
Solution:
Let’s take the first number as 20, which leaves a remainder of 8 when divided by 12.
Let’s take another number 32, which is 4 short of a multiple of 12.
Their sum = 20 + 32 = 52, which is not a multiple of 8.
So, Snehal claims is incorrect.

Question 3.
When is the sum of two multiples of 3, a multiple of 6, and when is it not? Explain the different possible cases, and generalise the pattern.
Solution:
Let the numbers be 3a and 3b.
Sum = 3a + 3b = 3(a + b)
Now, 3(a + b) is a multiple of 6 if both a and b are even or odd, so that their sum is also even.
And, 3(a + b) is not a multiple of 6 if one of them is odd and the other is even, so their sum is odd.
Thus, in general, the sum of two multiples of 3 is divisible by 6. If both the numbers are even or odd.
The sum of two multiples of 3 is not divisible by 6 if one number is even and the other is an odd multiple of 3.

Question 4.
Sreelatha says, “I have a number that is divisible by 9. If I reverse its digits, it will still be divisible by 9”.
(i) Examine if her conjecture is true for any multiple of 9.
(ii) Are any other digit shuffles possible such that the number formed is still a multiple of 9?
Solution:
(i) Consider the number 153
Now, 1 + 5 + 3 = 9, so it is divisible by 9.
If we reverse 153, we get 351.
Now, 3 + 5 + 1 = 9, which is also divisible by 9.
The sum of digits doesn’t change on reversing.
So, Sreelatha’s conjecture is true.

(ii) Yes, consider 972, where 9 + 7 + 2 = 18, which is divisible by 9.
If we shuffle 972, we get 279, 297, 729, 792, all have the same digital roots, which are divisible by 9.

Question 5.
If 48a23b is a multiple of 18, list all possible pairs of values for a and b.
Solution:
Consider 48a23b
Now, the given number is a multiple of 18 if it is divisible by both 2 and 9.
Now, 48a23b is divisible by 2 if ‘b’ = (0, 2, 4, 6, 8) and divisible by 9 if 4 + 8 + o + 2 + 3 + b = 17 + a + b is a multiple of 9.

So, all possible pairs of values for a and b are (1, 0), (8, 2), (6, 4), (4, 6), and (2, 8).

Question 6.
If 3p7q8 is divisible by 44, list all possible pairs of values for p and q.
Solution:
3p7q8 is divisible by 44 if it is divisible by both 4 and 11.
Now, a number is divisible by 4 if the number formed by the last two digits is divisible by 4.
∴ q = 0, 2, 4, 6, 8.
Now, the number is divisible by 11 if [(3 + 7 + 8) – (p + q)] is divisible by 11.

So, the possible pairs are (7, 0), (5, 2), (3, 4), and (1, 6).

Question 7.
Find three consecutive numbers such that the first number is a multiple of 2, the second number is a multiple of 3, and the third number is a multiple of 4. Are there more such numbers? How often do they occur?
Solution:
Yes, there are infinitely many such consecutive numbers, like 14, 15, 16; 26, 27, 28,…
∵ LCM (2, 3, 4) = 12
∴ New consecutive numbers will appear satisfying the given condition will appear at every 12th number.
i.e., 2 + 12 = 14, 3 + 12 = 15, 4 + 12 = 16,…
Thus, in general, the first number of this group can be described by the rule 12(n) + 2, as follows:
For n = 0, the first number for the group is 12 × 0 + 2 = 2
For n = 1, the first number for the group is 12 × 1 + 2 = 14
For n = 2, the first number for the group is 12 × 2 + 2 = 26, and so on.

Question 8.
Write five multiples of 36 between 45,000 and 47,000. Share your approach with the class.
Solution:
Consider 45,000 = 36(1250)
47,000 = 36(1305) + 20 = 46980 + 20
So, we have to find the multiples of 36 between 45,000 and 46980.
So, the five multiples of 36 are 45000, 45036, 45144, 45180, 45252.

Question 9.
The middle number in the sequence of 5 consecutive even numbers is 5p. Express the other four numbers in sequence in terms of p.
Solution:
Given the middle number in the sequence of 5 consecutive even numbers is 5p.
So, the other number are 5p – 4, 5p – 2, 5p + 2, 5p + 4.

Question 10.
Write a 6-digit number that is divisible by 15, such that when the digits are reversed, it is divisible by 6.
Solution:
A number is divisible by 15 if it is divisible by both 3 and 5.
i.e., the number ends in 0 or 5, and the sum of digits is divisible by 3.
Also, the reverse of that number is divisible by 6
i.e., it should be divisible by both 2 and 3.
Consider the number 630255, which is divisible by both 3 and 5.
The reverse of 630255 is 552036, which is divisible by both 2 and 3.
Thus, 630255 is divisible by 15, and when the digits are reversed
i.e., 552036 is divisible by 6.

Question 11.
Deepak claims, “There are some multiples of 11 which, when doubled, are still multiples of 11. But other multiples of 11 don’t remain multiples of 11 when doubled”. Examine if his conjecture is true; explain your conclusion.
Solution:
Multiples of 11 are:
11 × 1 = 11
11 × 2 = 22
11 × 3 = 33
11 × 4 = 44
Now, when we doubled the above multiples, we get,
2 × 11 = 22, 2 × 22 = 44, 2 × 33 = 66, 2 × 44 = 88
So, doubling of any multiples of 11 is still a multiple of 11.
Thus, Deepak’s claim is incorrect.

Question 12.
Determine whether the statements below are ‘Always True’, ‘Sometimes True’, or ‘Never True’. Explain your reasoning.
(i) The product of a multiple of 6 and a multiple of 3 is a multiple of 9.
(ii) The sum of three consecutive even numbers will be divisible by 6.
(iii) If abcdef is a multiple of 6, then badcef will be a multiple of 6.
(iv) 8(7b – 3) – 4(11b + 1) is a multiple of 12.
Solution:
(i) 6 × 3 = 18
i.e., the product of a multiple of 6 and a multiple of 3 is a multiple of 9.
Consider, 6 × 6 = 36, which is also true.
So, this statement is ‘always true’.

(ii) Let the three consecutive even numbers be 2n, 2n + 2, 2n + 4.
∴ Sum = 2n + 2n + 2 + 2n + 4 = 6n + 6 = 6(n + 1), which is divisible by 6.
So, the statement is ‘always true’.

(iii) Since abcdef is a multiple of 6,
i.e., the one’s digit is an even number and the sum of digits a + b + c + d + e + f is divisible by 3 or the digital root of abcdef is 3.
As per instruction, the new number badcef, i.e., the one’s digit is the same, an even, and the sum of digits is also divisible by 3 (whatever the place of digits).
So, it is also divisible by 6, or a multiple of 6.
For example, 123456 is divisible by both 2 and 3 and so by 6.
And after changing the place of the digit as instructed, 214356 is also divisible by both 2 and 3, so by 6 or a multiple of 6.

(iv) On simplifying 8(7b – 3) – 4(11b + 1), we have 56b – 24 – 44b – 4 = 12b – 28, which is not a multiple of 12.
So, the above statement is ‘never true’.

Question 13.
Choose any 3 numbers. When is their sum divisible by 3? Explore all possible cases and generalise.
Solution:
Let us choose any these numbers, we have to find when their sum is divisible by 3.
Case 1: All three numbers are divisible by 3.
Example: 3, 6, 9 → 3 + 6 + 9 = 18 (which is divisible by 3).
Case 2: All three numbers give the same remainder when divided by 3.
Example: 1, 4, 7 → Leaves Remainder 1
Sum = 1 + 4 + 7 = 12 (which is divisible by 3).
Example: 2, 5, 8 → Leaves Remainder 2
Sum = 2 + 5 + 8 = 15 (which is divisible by 3)
Case 3: One number leaves a remainder of 0. The second leaves 1 and the third leaves 2.
Example: 3, 4, 5 → Remainders are 0, 1, 2, respectively.
Sum = 3 + 4 + 5 = 12 (which is divisible by 3).
We observe that the sum of three numbers is divisible by 3 if the sum of their remainders is also divisible by 3.

Question 14.
Is the product of two consecutive integers always a multiple of 2? Why? What about the product of these consecutive integers? Is it always a multiple of 6? Why or why not? What can you say about the product of 4 consecutive integers? What about the product of five consecutive integers?
Solution:
Product of two consecutive integers (always has one even), so it is always a multiple of 2.
Product of three consecutive integers (One divisible by 2 and one by 3), so it is always a multiple of 6.
The product of four consecutive integers is always divisible by 24.
The product of five consecutive integers is always divisible by 120.

Question 15.
Solve the cryptarithms.
(i) EF × E = GGG
(ii) WOW × 5 = MEOW
Solution:

Question 16.
Which of the following Venn diagrams captures the relationship between the multiples of 4, 8, and 32?

Solution:
(iv) Every multiple of 32 is also a multiple of 8 and 4. (∵ 32 = 8 × 4)
Every multiple of 8 is also a multiple of 4 (∵ 8 = 4 × 2)
But not every multiple of 4 is a multiple of 8 or 32.
So, (iv) captures the relationship between the multiples of 4, 8, and 32.