Number Play Class 8 Solutions Ganita Prakash Maths Chapter 5

Solving questions with the help of Ganita Prakash Class 8 Solutions and Class 8 Maths Chapter 5 Number Play NCERT Solutions Question Answer improves confidence.

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?

Intext Questions

Question 1.
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.


Evaluate each expression and write the result next to it. Do you notice anything interesting? (Page 112)
Solution:
When four consecutive numbers are used with all possible combinations of ‘+’ and signs, the results will always be even. This is because, regardless of the sign placement, the sum will always involve adding and subtracting an equal number of consecutive numbers. Specifically, the sum of four consecutive numbers is always even, and the differences between them also result in even numbers when combined with plus and minus signs.
Example with 3, 4, 5, 6:

3 + 4 + 5 + 6 = 18
3 + 4 + 5 – 6 = 6
3 + 4 – 5 + 6 = 8
3 + 4 – 5 – 6 = -4
3 – 4 + 5 + 6 = 10
3 – 4 + 5 – 6 = -2
3 – 4 – 5 + 6 = 0
3 – 4 – 5 – 6 = -12
Observations:
All results are even numbers.
There are only eight possible combinations.
The results can be positive, negative, or zero.

Question 2.
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? (Page 113)
Solution:
Let us take numbers 5, 6, 7, 8:
Now,

5 + 6 + 7 + 8 = 26
5 + 6 + 7 – 8 = 10
5 + 6 – 7 + 8 = 12
5 + 6 – 7 – 8 = -4
5 – 6 + 7 + 8 = 14
5 – 6 + 7 – 8 = -2
5 – 6 – 7 + 8 = 0
5 – 6 – 7 – 8 = -16
Conclusion:
Regardless of the starting consecutive numbers, the same pattern of even results emerges when using all possible combinations of plus and minus signs. This is because the sum and difference of consecutive numbers will always result in an even number.

Intext Questions

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. (Page 115)

1. 2a + 2b
2. 3g + 5h
3. 4m + 2n
4. 2u – 4v
5. 13k – 5k
6. 6m – 3n
7. x² + 2
8. b² + 1
9. 4k × 3j
Solution:
1. Analyse the expression 2a + 2b
2a is even because it’s a multiple of 2.
2b is even because it’s a multiple of 2.
The sum of two even numbers is even, so 2a + 2b is always even.

2. Analyse the expression 3g + 5h
If g is odd and h is odd, then 3g is odd, and 5h is odd.
The sum of two odd numbers is even,
e.g., 3 × 1 + 5 × 1 = 8.
If g is even and h is even, then 3g and 5h are even.
The sum of two even numbers is even,
e.g., 3 × 2 + 5 × 2 = 16.
If g is odd and h is even, then 3g is odd, and 5h is even.
The sum of an odd and an even number is odd,
e.g., 3 × 1 + 5 × 2 = 13
Thus, 3g + 5h is not always even.

3. Analyse the expression 4m + 2n
4m is even because it’s a multiple of 2.
2n is even because it’s a multiple of 2.
The sum of two even numbers is even, so 4m + 2n is always even.

4. Analyse the expression 2u – 4v
2u is even because it’s a multiple of 2.
4v is even because it’s a multiple of 2.
The difference of two even numbers is even, so 2u – 4v is always even.

5. Analyse the expression 13k – 5k
This simplifies to 8k.
8k is even because it’s a multiple of 2.
Thus, 13k – 5k is always even.

6. Analyse the expression 6m – 3n
6m is always even.
3n can be odd (if n is odd) or even (if n is even).
If n is odd, 3n is odd, and the difference of an even and an odd number is odd,
e.g. 6 × 1 – 3 × 1 = 3.
Thus, 6m – 3n is not always even.

7. Analyse the expression x² + 2
x² is odd or even, both.
If x² is even, x² + 2 becomes even.
If x² is odd, x² + 2 becomes odd.
Thus, x² + 2 is not always even.

8. If b is even, b2 is even, and b² + 1 odd,
e.g., 2² + 1 = 5.
If b is odd, b² is odd, and b² + 1 is even,
e.g, 3² + 1 = 10.
Thus, b² + 1 is not always even.

9. Analyse the expression 4k × 3j
This simplifies to 12kj.
12kj is even because it’s a multiple of 2.
Thus, 4k + 3j is always even.

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 that 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:
Since p is the greatest number. So, the numbers before p will be p – 1, p – 2, p – 3, and p – 4.
Therefore, the other four numbers are p – 1, p – 2, p – 3, and p – 4.

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) Sometimes true: the sum of two even numbers is a multiple of 3.
Examples:

• 2 + 4 = 6, 8 + 10 = 18, 14 + 16 = 30 are multiples of 3.
• 2 + 6 = 8, 4 + 10 = 14 are not multiples of 3.

(ii) Sometimes true: if a number is not divisible by 18, then it is also not divisible by 9.
Examples:

• 27 is not divisible by 18, but 27 is divisible by 9. True
• 40 is not divisible by 18, also it is not divisible by 9. False

(iii) Never true, if two numbers are not divisible by 6, then their sum is not divisible by 6.
Examples:

• 8 and 10 are not divisible by 6.
  The sum of two numbers = 8 + 10 = 18, is divisible by 6.
• 10 and 13 are not divisible by 6.
  The sum of 10 and 13 = 10 + 13 = 23, which is not divisible by 6.

(iv) Always true
Multiple of 6; 6m
Multiple of 9; 9n
6m + 9n = 3(2m + 3n)
Hence multiple of 3.

(v) Sometimes true, the sum of a multiple of 6 and a multiple of 3 is a multiple of 9.
Multiples of 6 are: 6, 12, 18, 24, 30,…
6 + 12 = 18 is a multiple of 9.
12 + 18 = 30 is not a multiple of 9.
18 + 24 = 42 is not a multiple of 9.
Sometimes true.
Multiples of 3 are: 3, 6, 9, 12, 15, 18,…
3 + 6 = 9 is a multiple of 9.
6 + 9 = 15 is not a multiple of 9.
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:
Let x be the number.
When x is divided by 3, the remainder is 2.
∴ x = 3k + 2, where k is any whole number.
If k = 0, x = 3 × 0 + 2 = 2
If k = 1, x = 3 × 1 + 2 = 5
If k = 2, x = 3 × 2 + 2 = 8
If k = 3, x = 3 × 3 + 2 = 11
If k = 4, x = 3 × 4 + 2 = 14
And so on.
When x is divided by 4, the remainder is 2.
x = 4m + 2, where m is any whole number.
If m = 0, then x = 4 × 0 + 2 = 2
If m = 1, then x = 4 × 1 + 2 = 6
If m = 2, then x = 4 × 2 + 2 = 10
If m = 3, then x = 4 × 3 + 2 = 14
If m = 4, then x = 4 × 4 + 2 = 18 And so on.
Now, the numbers that appear in both lists are 2 and 14. It we continue the lists, the next one would be 26.
Now, the difference between these numbers are 14 – 2 = 12 and 26 – 14 = 12 The difference between consecutive numbers is 12 and 12 is the LCM of the divisors 3 and 4.
We have the common numbers are 2, 14, 26
Here, each number is 12 more than the previous one.
All these numbers leave a remainder of 2 when divided by both 3 and 4.
Required algebraic expression is x = LCM (3 and 4) n + 2
= 12n + 2, where n 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:
The LCM of 3, 5, and 7 = 3 × 5 × 7 = 105 [∵ 3, 5, and 7 are prime numbers]
No. of pebbles = 105 + 1 = 106

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 77? Show the solution both algebraically and visually.
(i) 4779 + 661
(ii) 4779 – 661
Solution:
Given, 661 = K × 7 + 3, where K = 1, 2, 3, 4, …
and, 4779 = K × 7 + 5
Algebraic Method:
(i) 4779 + 661
4779 = (682 × 7 ) + 5
Remainder = 5
661 = (94 × 7) + 3
Remainder = 3
∴ \(\frac{4779+661}{7}=\frac{5+3}{7}=\frac{8}{7}\) = 1 = Remainder
(ii) 4779 – 661
∴ \(\frac{4779-661}{7}=\frac{5-1}{7}=\frac{8}{7}\) = 4 = Remainder

Visualization Method:
(i) 4779 + 661 = (682 × 7) + 5 + (94 × 7) + 3
= 7 × (682 + 94) + 5 + 3
= 7 × 776 + 8
= Divisible by 7 + \(\frac {8}{7}\)
= 1, Remainder
(ii) 4779 – 661 = (682 × 7) + 5 – (94 × 7) – 3
= 7 × (682 – 94) + 5 – 3
= 7 × 588 + 2
= Divisible by 7 + 2
= 2, Remainder

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:
When divided by 3, the remainder is 2.
∴ The number is the form 3k + 2.
When divided by 4, the remainder is 3.
∴ The number is the form 4m + 3.
When divided by 5, the remainder is 4.
∴ The number is the form 5n + 4.
Here, in each case, the remainder is one less than the divisor,
i.e. 3 – 2 = 1, 4 – 3 = 1, 5 – 4 = 1
If we add 1 to the number, it will be perfectly divisible by 3, 4 and 5.
Since adding 1 to the number makes it divisible by 3, 4 and 5, the remainder plus 1 must be a multiple of the LCM of 3, 4, and 5.
LCM (3, 4, 5) = 3 × 4 × 5 = 60
∴ Required number = 60 – 1 = 59
Explanation of why it is the smallest.
The property of the remainders being less than the respective divisors means that if we add 1 to the number, it becomes a multiple of all three divisors.
To find the smallest such number, we need the smallest multiple of all three divisors, which is their LCM.
Since 60 is the smallest positive common multiple of 3, 4, and 5, subtracting 1 from it gives the smallest number satisfying the condition.

5.2 Checking Divisibility Quickly

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:
If the sum of the digits of a number is divisible by 9, then the number is divisible by 9.
(i) Sum of the digits = 1 + 2 + 3 = 6, is not divisible by 9.
Thus, 123 is not divisible by 9.

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

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

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

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

Question 2.
Find the smallest multiple of 9 with no odd digits.
Solution:
To find the smallest multiple of 9 with no odd digits,
we can list multiples of 9 and check their digits.
We need a number that is a multiple of 9 and all its digits must be even i.e. 0, 2, 4, 6, 8.
So, multiples of 9 are 9 contains odd digit 9,
18 contains odd digit 1,
27 contains odd digit 7,
36 contains odd digit 3,
.
.
.
81 contains odd digit 1,
90 contains odd digit 9,
.
.
.
108 contains odd digit 1.
288 has all digits even.
Sum of digits = 2 + 8 + 8 = 18, which is a multiple of 9.
So, the smallest multiple of 9 with no odd digits is 288.

Question 3.
Find the multiple of 9 that is closest to the number 6000.
Solution:
Given, 6000
Sum of the digits = 6 + 0 + 0 + 0 = 6
We know that, if the number is divisible by 9, then the sum of the digits is divisible by 9.
If we add 3 to the number 6000.
6000 + 3 = 6003, it is divisible by 3.
Thus, the multiple of 9 that is closest to the number is 6003.

Question 4.
How many multiples of 9 are there between the numbers 4300 and 4400?
Solution:
The multiples of 9 are there between the numbers 4300 and 4400 are 4302, 4311, 4320,………, 4392
The number of multiples of 9 = \(\frac{\text { Last term }- \text { First term }}{\text { Difference }}+1\)
= \(\frac{4392-4302}{9}+1\)
= \(\frac {90}{9}\) + 1
= 10 + 1
= 11
Thus, the multiples of 9 are 11.

Intext Questions (Pages 130)

Question 1.
Between the numbers 600 and 700, which numbers have the digital root:
(i) 5
(ii) 7
(iii) 3
Solution:
(i) Digital root 5:
608 = 6 + 0 + 8 = 14 = 1 + 4 = 5;
617 = 6 + 1 + 7 = 14 = 1 + 4 = 5;
662 = 6 + 6 + 2 = 14 = 1 + 4 = 5;
689 = 6 + 8 + 9 = 23 = 2 + 3 = 5, etc.

(ii) Digital root 7:
610 = 6 + 1 + 0 = 7;
619 = 6 + 1 + 9 = 16 = 1 + 6 = 7;
637 = 6 + 3 + 7 = 16 = 1 + 6 = 7;
673 = 6 + 7 + 3 = 16 = 1 + 6 = 7, etc.

(iii) Digital root 3:
606 = 6 + 0 + 6 = 12 = 1 + 2 = 3;
615 = 6 + 1 + 5 = 12 = 1 + 2 = 3;
633 = 6 + 3 + 3 = 12 = 1 + 2 = 3;
678 = 6 + 7 + 8 = 21 = 2 + 1 = 3, etc.

Question 2.
Write the digital roots of any 12 consecutive numbers. What do you observe?
Solution:
The digital roots of any 12 consecutive numbers are:

• 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;
• 113 = 1 + 1 + 3 = 5;
• 114 = 1 + 1 + 4 = 6;
• 115 = 1 + 1 + 5 = 7;
• 116 = 1 + 1 + 6 = 8

Observation:
The digital roots of cycles repeat after 9 numbers.
6 → 7 → 8 → 9 → 1 → 2 → 3 → 4 → 5 → 6 → 7 → 8
So, the digital roots of consecutive numbers form a repeating cycle of length 9.
The digital root of multiples by 9:

• 405 = 4 + 0 + 5 = 9;
• 234 = 2 + 3 + 4 = 9;
• 1035 = 1 + 0 + 3 + 5 = 9;
• 936 = 9 + 3 + 6 = 18 = 1 + 8 = 9, etc.

Question 3.
We saw that the digital root of multiples by 9 is always 9. Now, find the digital roots of some consecutive multiples of (i) 3, (ii) 4, and (iii) 6.
Solution:
(i) The digital roots of some consecutive multiples of 3 are:
39 = 3 + 9 = 12 = 1 + 2 = 3;
42 = 4 + 2 = 6;
45 = 4 + 5 = 9;
48 = 4 + 8 = 12 = 1 + 2 = 3;
51 = 5 + 1 = 6;
54 = 5 + 4 = 9;
57 = 5 + 7 = 12 = 1 + 2 = 3;
60 = 6 + 0 = 6;
63 = 6 + 3 = 9 etc.
Thus, the digital roots of consecutive multiples of 3 are 3, 6, 9, 3, 6, 9,……

(ii) The digital roots of some consecutive multiples of 4 are:
32 = 3 + 2 = 5;
36 = 3 + 6 = 9;
40 = 4 + 0 = 4;
44 = 4 + 4 = 8;
48 = 4 + 8 = 12 = 1 + 2 = 3;
52 = 5 + 2 = 7;
56 = 5 + 6 = 11 = 1 + 1 = 2;
60 = 6 + 0 = 6, etc.
Thus, the digital roots of consecutive multiples of 4 are 5, 9, 4, 8, 3, 7, 2, 6,……..

(iii) The digital roots of some consecutive multiples of 6 are:
30 = 3 + 0 = 3;
36 = 3 + 6 = 9;
42 = 4 + 2 = 6;
48 = 4 + 8 = 12 = 1 + 2 = 3;
54 = 5 + 4 = 9;
60 = 6 + 0 = 6;
66 = 6 + 6 = 12 = 1 + 2 = 3;
72 = 7 + 2 = 9;
78 = 7 + 8 = 15 = 1 + 5 = 6, etc.
Thus, the digital roots of consecutive multiples of 6 are 3, 9, 6, 3, 9, 6, 3, 9, 6,……….

Question 4.
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.
Solution:
The digital roots of the numbers that are 1 more than a multiple of 6 are:
37 = 3 + 7 = 10 = 1 + 0 = 1;
43 = 4 + 3 = 7;
49 = 4 + 9 = 13 = 1 + 3 = 4;
55 = 5 + 5 = 10 = 1 + 0 = 1;
61 = 6 + 1 = 7;
67 = 6 + 7 = 13 = 1 + 3 = 4, etc.
Hence, the digital roots of the numbers that are 1 more than a multiple of 6 are 1, 7, 4, 1, 7, 4,…..
We notice the digital roots cycle through 1, 7, 4, and then repeat: 1, 7, 4, 1, 7, 4, 1, 7, 4,………

Question 5.
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:
Try: 111 111 111
Digits = 9
Sum = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 9
Digital root = 9
All digits are odd (1)
Satisfies all the conditions.
Hence, the answer is 111 111 111.

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 number = 40
The sequence of numbers by repeatedly adding 11 are 40, 51(40 + 11), 62(51 + 11), 73(62 + 11), 84(73 + 11), 95(84 + 11), 106(95 + 11), 117(106 + 11), 128(117 + 11), 139(128 + 11), etc.
The digital roots of this sequence of numbers are:
40 = 4 + 0 = 4;
51 = 5 + 1 = 6;
62 = 6 + 2 = 8;
73 = 7 + 3 = 10 = 1 + 0 = 1;
84 = 8 + 4 = 12 = 1 + 2 = 3;
95 = 9 + 5 = 14 = 1 + 4 = 5;
106 = 1 + 0 + 6 = 7;
117 = 1 + 1 + 7 = 9;
128 = 1 + 2 + 8 = 11 = 1 + 1 = 2;
139 = 1 + 3 + 9 = 13 = 1 + 3 = 4,… etc.
Thus, the digital roots of this sequence of numbers are 4, 6, 8, 1, 3, 5, 7, 9, 2, 4,…..
Observations:
The digital roots are 4, 6, 8, 1, 3, 5, 7, 9, 2, 4,……
This sequence starts repeating after 9 steps.
So the digital roots form a cycle: 4, 6, 8, 1, 3, 5, 7, 9, 2, 4,……

Question 3.
What will be the digital root of the number 9a + 36b + 13?
Solution:
We have, 9a + 36b + 13
Since 9a is a multiple of 9.
∴ The digital root of 9a = 9
Also, 36b is a multiple of 9.
∴ Digital root of 36b = 9
Now, digital root of 13 = 1 + 3 = 4
∴ Digital root of 9a + 36
b + 13 = 9 + 9 + 4 = 22
and digital root of 22 = 2 + 2 = 4
∴ Digital root of 9a + 36b + 13 = 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:
Consider the pattern: 8, 16, 24, 32, 40,……
(i) 8 = 8 = digital root, parity → even
16 = 1 + 6 = 7 = digital root, parity → odd
24 = 2 + 4 = 6 = digital root, parity → even
32 = 3 + 2 = 5 = digital root, parity → odd
40 = 4 + 0 = 4 = digital root, parity → even

(ii) Divided by 3
8 ÷ 3 ⇒ 2, Remainder
24 ÷ 3 ⇒ 0, Remainder
32 ÷ 3 ⇒ 2, Remainder
40 ÷ 3 ⇒ 1, Remainder
Divided by 9
8 ÷ 9 ⇒ 8, Remainder
24 ÷ 9 ⇒ 6, Remainder
32 ÷ 9 ⇒ 5, Remainder
40 ÷ 9 ⇒ 4, Remainder

5.3 Digits in Disguise

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 12.1 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:
A number that leaves a remainder of 8.
when divided by 12: 12k + 8, where k ≥ 1.
Also, another number 4 short of a multiple of 12: 12k – 4

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:
Multiples of 3 are: 3, 6, 9, 12, 15, 18,……….
3 + 6 = 9, not a multiple of 6.
6 + 9 = 15, not a multiple of 6.
3 + 9 = 12, multiple of 6.
6 + 12 = 18, multiple of 6.
There are two possible cases.

• If both numbers are odd, then the sum is a multiple of 6.
• If both numbers are even, then the sum is a multiple of 6.

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:
Consider a number that is divisible by 9 = 72
We know that,
If the sum of the digits is divisible by 9, then the number is divisible by 9.
If its digits are reversed
27 = 2 + 7 = 9, it is also divisible by 9.
(i) True
(ii) Yes, any other digit shuffle is possible, and the number is still a multiple of 9.

Question 5.
If 48a23b is a multiple of 18, list all possible pairs of values for a and b.
Solution:
We know that, if a number is a multiple of 18 then it is multiple of both 2 and 9.
For 48a23h to be a multiple of 2, the last digit b must be an even number.
So, b must be 0, 2, 4, 6, 8.
For 48a23h to be a multiple of 9, the sum of its digits must be a multiple of 9.
Sum of digits of 48a23b = 4 + 8 + a+ 2+ 3 + b
= 17+a + b …(i)
So, 17 + a + b must be a multiple of 9.
If b = 0, then 17+a + 0=17+o, [from Eq. (i)]
which is a multiple of 9.
So, a must be 1.
∴ (a,b) = (1, 0)
If b = 2, then 17 + a + 2 = 19 + a, which is a multiple of 9. So, a must be 8.
∴ (a, b) = (8, 1)
If b = 4, then 17 + a + 4 = 21 + a, which is a multiple of 9. So, a must be 6.
(a, b) = (6, 4)
If b = 6, then 17 + a + 6 = 23 + a, which is a multiple of 9. So, a must be 4.
∴(a, b) = (4, 6)
If = 8, then 17 + a + 8 = 25 + a, which is a multiple of 9. So, a must be 2.
∴(a, b) = (2, 8)

Question 6.
If 3p7q8 is divisible by 44, list all possible pairs of values for p and q.
Solution:
Given by question, 3p7q8 is divisible by 44.
As we know, if a number is divisible by 44, then it is also divisible by 4 and 11.
∴ 3p7q8
Case 1: Put p = 1 and q = 0
37708 is divisible by 4 and 11, then it is also divisible by 44.

Case 2: Put p = 5 and q = 2
35728 is divisible by 4 and 11, then it is also divisible by 44.

Case 3: Put p = 3 and q = 4
33748 is divisible by 4 and 11, then it is also divisible by 44.

Case 4: Put p = 1 and q = 6
31768 is divisible by 4 and 11, then it is also divisible by 11.
Thus, (p = 7, q = 0), (p = 5, q = 2), (p = 3, q = 4), and (p = 1 and q = 6) are the possible pairs of values for p and q.

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 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 the sequence in terms of p.
Solution:
Given the middle number in the sequence of 5 consecutive even numbers 5p.
The other four numbers in the sequence in terms of p are 5p – 4, 5p – 2, 5p + 2, 5p + 4
Hence, the other four numbers in the sequence are p, 3p, 7p, and 9p.

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:
We know that, a number is divisible by 15 if it is divisible by both 3 and 5 and a number is divisible by 6 if it is divisible both 2 and 3.
Let 6-digit number be abcdef.
For abcdef to be divisible by 15, f must be 0 or 5 and (a + b + c + d + e + f) must be divisible by 3.
The reversed number is fedcba.
For fedcba to be divisible by 6, a must be even and (f + e + d + c + b + a) must be divisible by 3.
Since, a must be even (for the reversed number to be divisible by 2) and f must be 0 or 5 (for the original number to be divisible by 5), we can choose values that satisfy these.
Ths sum of digits (a + b + c + d + e + f) being divisible by 3 is required for both the original and reversed number when is consistent.

Let’s choose f = 0 (to satisfy divisibility by 5 for the original number).
Let’s choose a = 2 (to satisfy divisibility by 2 for the reversed number).
Now we need to place b, c, d, e such that the sum of digits (2 + b + c + d + e + 0) is divisible by 3.
Let’s pick simple digits to make the sum divisible by 3.
For example, Let b = 1, c = 1, d = 1, e = 1
The-sum of digits would be 2 + 1 + 1 + 1 + 1 + 0 = 6, which is divisible by 6.
So, the original number is 211110.
Reversed number is 011112 i.e. 11112
∴ Required 6-digit number = 211110

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:
The multiples of 11 are: 11, 22, 33, 44, 55,…
When doubled, 22, 44, 66, 88, 110,……
i.e. (11) × 2, 11 × 4, 11 × 6, 11 × 8, 11 × 10,…. are also multiples of 11.
False, if multiples of 11 are doubled, then the multiples of 11 are these numbers.

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) Always True,
The multiple of 6 can be written as 6a, where a is an integer.
The multiple of 3 can be written as 3b, where b is an integer.
∴ Product = (6a) × (3b) = 18(ab) is a multiple of 9.

(ii) Always True,
The sum of three consecutive even numbers will be divisible by 6.
For example 2 + 4 + 6 = 12, 4 + 6 + 8 = 18, 6 + 8 + 10 = 24, 8 + 10 + 12 = 30,…
These numbers are divisible by 6.

(iii) Always True, because one’s place does not change.

(iv) Sometimes true,
Conclusion:
8(7 × 1 – 3) – 4(11 × 1 + 1) = -16, not divisible by 12.
8(7 × 10 – 3) – 4(4 × 10 + 1) = 536 – 164 = 372, divisible by 12.

Question 13.
Choose any 3 numbers. When is their sum divisible by 3? Explore all possible cases and generalise.
Solution:
Let the three numbers be n₁, n₂, and n₃.
Let their remainders when divided by 3 be r₁, r₂, and r₃.
The sum n₁ + n₂ + n₃ is divisible by 3 if and only if r₁ + r₂ + r₃ is divisible by 3.
Case 1: All remainders are 0.
r₁ = 0, r₂ = 0, r₃ = 0
Sum of remainders = 0 + 0 + 0 = 0, which is divisible by 3.

Case 2: All remainders are 1.
r₁ = 1, r₂ = 1, r₃ = 1
Sum of remainders = 1 + 1 + 1 = 3, which is divisible by 3.

Case 3: All remainders are 2.
r₁ = 2, r₂ = 2, r₃ = 2
Sum of remainders = 2 + 2 + 2 = 6, which is divisible by 3.

Case 4: One remainder is 0, one is 1, and one is 2.
r₁ = 0, r₂ = 1, r₃ = 2 (in any order).
Sum of remainders = 0 + 1 + 2 = 3, which is divisible by 3.
The sum of three numbers is divisible by 3 if and only if all three numbers have the same remainder when divided by 3, or if they all have different remainders when divided 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:
Yes, the product of two consecutive integers is always a multiple of 2. In any pair of consecutive integers, one integer must be even, and the other must be odd.
When an even number is multiplied by any other integer, the result is always even, and it is divisible by 2.
Yes, the product of three consecutive integers is always a multiple of 6. Among three consecutive integers, at least one is a multiple of 2 and exactly one is a multiple of 3.
Since the product contains a factor of 2 and a factor of 3. So, it must be a multiple of 6.
The product of 4 consecutive integers is always a multiple of 24. Among any four consecutive integers, at least two are even numbers, one of which is a multiple of 4.
So, the product is a multiple of 8.
And at least one is a multiple of 3.
Since the product contains factors of 8 and 3.
So, it must be a multiple of their LCM, which is 24.
The product of five consecutive integers is always a multiple of 120.
Among any five consecutive numbers, at least two are even numbers, one of which is a multiple of 4, atleast one is a multiple of 3 and exactly one is a multiple of 5.
So, the product is a multiple of 2 × 3 × 4 × 5 = 120

Question 15.
Solve the cryptarithms
(i) EF × E = GGG
(ii) WOW × 5 = MEOW
Solution:
(i) This means a 2-digit number multiplied by 5 gives a 3-digit number.
2-digit number = 20, 21,…., 99
37 × 3 = 111; all conditions are satisfied.

(ii) This means a 3-digit number multiplied by 5 gives 4-digit numbers.
Pick 3-digit number = 200, 201,…., 999
525 × 5 = 2625

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

Solution:
(iv) Multiples of 4 are: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64,…
Multiples of 8 are: 8, 16, 24, 32, 40, 48, 56, 64,….
Multiples of 32 are: 32, 64, 96, 128,…
The Venn diagram captures the relationship between the multiples of 4, 8, and 32: