Playing with Numbers
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NCERT Chapter Answers and other Q & As
Q1: Write the following numbers in generalised form.
(i) 25 (ii) 73 (iii) 129 (iv) 302
Answer:
(i) 25 = 10 × 2 + 5 (ab = 10 ×a + b)
(ii) 73 = 10 × 7 + 3
(iii) 129 = 100 × 1 + 10 × 2 + 9 (abc = 100 × a + 10 × b + c)
(iv) 302 = 100 × 3 + 10 × 0 + 2
Q2: Write the following in the usual form.
(i) 10 × 5 + 6
(ii) 100 × 7 + 10 × 1 + 8
(iii) 100 × a + 10 × c + b
Answer:
(i) 10 × 5 + 6 = 50 + 6 = 56
(ii) 100 × 7 + 10 × 1 + 8
(iii) 100 × a + 10 × c + b
Q3: Find A and B in the addition.
A + A + A ------ B A ------
Answer:
5 + 5 + 5 ------ 1 5 ------
NCERT EXERCISE 16.1
Q4: Find the values of the letters in each of the following and give reasons for the steps involved.
1. 3 A 2. 4 A 3. 1 A
+2 5 +9 8 × A
----- ------ -----
B 2 C B 3 9 A
----- ------ -----
4. A B 5. A B 6. A B
+ 3 7 × 3 × 5
------ ------ -------
6 A C A B C A B
------ ------ -------
7. A B 8. A 1 9. 2 A B
× 6 +1 B +A B 1
------ ------ -------
B B B B 0 B 1 8
------ ------ -------
10. 1 2 A
+ 6 A B
--------
A 0 9
--------
Answer:
1. 3 7 2. 4 5 3. 1 6
+2 5 +9 8 × 6
----- ------ -----
6 2 1 4 3 9 6
----- ------ -----
4. 2 5 5. 5 0 6. 5 0
+ 3 7 × 3 × 5
------ ------ -------
6 2 1 5 0 2 5 0
------ ------ -------
7. 7 4 8. 7 1 9. 2 4 7
× 6 +1 9 +4 7 1
------ ------ -------
4 4 4 9 0 7 1 8
------ ------ -------
10. 1 2 8
+ 6 8 1
--------
8 0 9
--------
See the article on ☛Rules of Divisibility.
Q5: Check the divisibility of the following numbers by 9.
(i) 108
(ii) 616
(iii) 294
(iv) 432
(v) 927
Answer:
(i) 108 = 1 + 0 + 8 = 9 (multiple of 9), ∴ Number is divisible by 9.
(ii) 616 = 6 + 6 + 1 = 13 (not multiple of 9), ∴ Number is not divisible by 9.
(iii) 294 = 2 + 9 + 4 = 15 (not multiple of 9), ∴ Number is not divisible by 9.
(iv) 432 = 4 + 3 + 2 = 9 (multiple of 9), ∴ Number is divisible by 9.
(v) 927 = 9 + 2 + 7 = 18 (multiple of 9), ∴ Number is divisible by 9.
Q6: You have seen that a number 450 is divisible by 10. It is also divisible by 2 and 5 which are factors of 10. Similarly, a number 135 is divisible 9. It is also divisible by 3 which is a factor of 9. Can you say that if a number is divisible by any number m, then it will also be divisible by each of the factors of m?
Answer: Yes. If a number is divisible by any number say 'm', then it is divisible by factors of m also.
450 = 10 × 45 = 5 × 2 × 45
Factors of 10 are 2,5 ⇒ 450 is also divisible by factors of 10.
Q7: According to the divisibility rule of 3, sum of digits must be equal to 3. Why does this rule work? Prove it for a two digit number ab.
Answer: Let a 2-digit number = ab
The number can be represented as = 10a + b
= 9a + a + b
= 9a + (a + b)
Since 9a is already divisible by 3, it implies (a + b) must be divisible by 3.
∴ we can infer the sum of digits must be multiple of 3 to test if a number can be divided completely by 3.
Q8: (i) Write a 3-digit number abc as 100a + 10b + c
= 99a + 11b + (a – b + c)
= 11(9a + b) + (a – b + c)
If the number abc is divisible by 11, then what can you say about (a – b + c)?
Is it necessary that (a + c – b) should be divisible by 11?
(ii) Write a 4-digit number abcd as 1000a + 100b + 10c + d
= (1001a + 99b + 11c) – (a – b + c – d)
= 11(91a + 9b + c) + [(b + d) – (a + c)]
If the number abcd is divisible by 11, then what can you say about [(b + d) – (a + c)]?
(iii) From (i) and (ii) above, can you say that a number will be divisible by 11 if the difference between the sum of digits at its odd places and that of digits at the even places is divisible by 11?
Answer: (i) the number consists of two parts i.e. 11(9a + b) and (a - b + c)
Since the part 11(9a + b) is already a multiple of 11 and hence it divisible by 11.
To make the number abc be divisible by the second part ( a - b + c) must be divisible by 11.
(ii) 4-digit number abcd is sum of two parts i.e. 11(91a + 9b + c) and [(b + d) – (a + c)]
The first part 11(91a + 9b + c) is already a multiple of 11. For the 4-digit number be divisible by 11, second part i.e. [(b + d) – (a + c)] must be divisible by 11.
(iii) From (i) and (ii) we infer that (a – b + c) and [(b + d) – (a + c)] must be multiple of 11. We can say any number will be divisible by 11 if the difference between the sum of digits at its odd places and that of digits at the even places is divisible by 11.
NCERT EXERCISE 16.2
Q9: If 21y5 is a multiple of 9, where y is a digit, what is the value of y?
Answer: If a number is a multiple of 9 then sum of its digits must also be multiple of 9
⇒ 2 + 1 + y + 5 = y + 8
⇒ y must be 1 to make the sum 9.
∴ y = 1
Q10: If 31z5 is a multiple of 9, where z is a digit, what is the value of z?
You will find that there are two answers for the last problem. Why is this so?
Answer: If a number is a multiple of 9 then sum of its digits must also be multiple of 9
⇒ 3 + 1 + z + 5 = 9 + z
⇒ z must be either 0 or 9 to make the entire sum multiple of 9.
∴ z = 0 or 9
Q11: If 24x is a multiple of 3, where x is a digit, what is the value of x?
Answer: The sum of digits of the 3-digit number must be 3
i.e. 2 + 4 + x = 6 + x.
Since x is a single digit which can values from 0 to 9. Possible values of x such that sum (6 + x ) remains multiple of 3 will be
6 + 0 = 6,
6 + 3 = 9,
6 + 6 = 12
6 + 9 = 15.
∴ x can have four possible values i.e. 0, 3,6 and 9.
Q12: If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?
Answer: The sum of 4 digits must be multiple of 3 i.e.
3 + 1 + z + 5 = 9 + z
Following values of z are possible so that sum (9 + z) must be multiple of 3
9 + 0 = 9
9 + 3 = 12
9 + 6 = 15
9 + 9 = 18
∴ z can have four possible values i.e. 0, 3,6 and 9.
Q13: Find the possible values of letters in the following:
1 A 3 B 5
- 5 C 7 D
-------------
6 6 6 6
-------------
Answer:
1 2 3 4 5
- 5 6 7 8
-------------
6 6 6 6
-------------
Q 14: Take any three-digit integer, reverse its digits, and subtract. For example, 742 − 247 = 495. The difference is divisible by 9. Prove that this must happen for all three-digit numbers abc.
Answer: Let the number = abc = 100a + 10b + c
Reversed number will be = cba = 100c + 10b + a
Difference of the two numbers = abc - cba = (100a + 10b + c) - (100c + 10b + a)
= 100a - a + 10b - b + c - 100c
= 99a - 99c = 99(a -c)
= 9 × 11 × (a - c)
⇒ The difference will always be multiple of 9 and 11.
Q15: Identify the pattern and find the next three terms
(a) 3, 9, 15, 21, _, _, _
(b) 1, 4, 9, 16, ___ , ___ , ___ .
(c)1, 1, 2, 3, 5, 8, 13, 21, ___ , ___ , ___ .
Answer:
(a) 3, 9, 15, 21, __, __, __
Each number is 6 more than the previous number.
The next three terms are 27(21+6), 33(27+6), 39(33+6).
(b) 1, 4, 9, 16, ___ , ___ , ___ .
= 12, 22, 32 ...
The number is square of natural numbers.The next three terms are: 25, 36 and 49.
(c)1, 1, 2, 3, 5, 8, 13, 21, ___ , ___ , ___ .
The next number is sum of previous two numbers.
The next three terms are: 34, 55, 89
This series is called Fibonacci Series.
thanks
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