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PLAYING WITH NUMBERS
Exercise 3.1
Q.1. Write all the factors of the following numbers :
(a) 24 (b) 15
(c) 21
(d) 27
(e) 12
(f) 20 (g) 18
(h) 23
(i) 36
Ans. (a) 24 = 1 × 24
= 2 × 12
= 3×8
= 4×6
Hence, factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24.
(b) 15 = 1 × 15
= 3×5
Hence, factors of 15 are 1, 3, 5 and 15.
(c) 21 = 1 × 21
= 3×7
Hence, factors of 21 are 1, 3, 7 and 21.
(d) 27 = 1 × 27
= 3×9
Hence, factors of 27 are 1, 3, 9 and 27.
(e) 12 = 1 × 12
= 2×6
= 3×4
Hence, factors of 12 are 1, 2, 3, 4, 6 and 12.
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(f) 20 = 1 × 20
= 2 × 10
= 4×5
Hence, factors of 20 are 1, 2, 4, 5 , 10 and 20.
(g) 18 = 1 × 18
= 2×9
= 3×6
Hence, factors of 18 are 1, 2, 3, 6, 9 and 18.
(h) 23 = 1 × 23
Hence, factors of 23 are 1, and 23.
(i) 36 = 1 × 36
= 2 × 18
= 3 × 12
= 4×9
= 6×6
Hence, factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36.
Q.2. Write first five multiples of :
(a) 5
(b) 8
(c) 9
Ans. (a) The required multiples of 5 are :
5 × 1 = 5, 5 × 2 = 10, 5 × 3 = 15, 5 × 4 = 20, 5 × 5 = 25.
i.e., 5, 10, 15, 20 and 25
(b) The required multiples of 8 are :
8 × 1 = 8, 8 × 2 = 16, 8 × 3 = 24, 8 × 4 = 32, 8 × 5 = 40.
i.e., 8, 16, 24, 32 and 40
(c) The required multiples of 9 are :
9 × 1 = 9, 9 × 2 = 18, 9 × 3 = 27, 9 × 4 = 36, 9 × 5 = 45.
i.e., 9, 18, 27, 36 and 45
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Q.3. Match the items in column 1 with the items in column 2.
Column I
(i) 35
(ii) 15
(iii) 16
(iv) 20
(v) 25
Column 2
(a) Multiple of 8
(b) Multiple of 7
(c) Multiple of 70
(d) Factor of 30
(e) Factor of 50
(f) Factor of 20
Ans. (i) b, (ii) d, (iii) a, (iv) f, (v) e
Q.4. Find all the multiples of 9 upto 100.
Ans. 9 × 1 = 9, 9 × 2 = 18, 9 × 3 = 27, 9 × 4 = 36, 9 × 5 = 45,
9 × 6 = 54, 9 × 7 = 63, 9 × 8 = 72, 9 × 9 = 81,
9 × 10 = 90, 9 × 11 = 99 and next multiple number is
greater than 100. So, all the multiples of 9 upto 100 are 9,
18, 27, 36, 45, 54, 63, 72, 81, 90 and 99.
Exercise 3.2
Q.1. What is the sum of any two (a) Odd numbers? (b) Even
numbers?
Ans. (a) The sum of any two odd numbers is even.
i.e. (1 + 3 = 4)
(b) Sum of any two even numbers is also even.
i.e. (6 + 2 = 8)
Q.2. State whether the following statements are True or
False:
(a) The sum of three odd numbers is even.
(b) The sum of two odd numbers and one even number
is even.
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(c) The product of three odd numbers is odd.
(d) If an even number is divided by 2, the quotient is
always odd.
(e) All prime numbers are odd.
(f) Prime numbers do not have any factors.
(g) Sum of two prime numbers is always even.
(h) 2 is the only even prime number.
(i) All even numbers are composite numbers.
(j) The product of two even numbers is always even.
Ans. (a) F
(b) T
(c) T
(d) F
(e) F
(f) F
(g) F
(h) T
(i) F
(j) T
Q.3. The numbers 13 and 31 are prime numbers. Both these
numbers have same digits 1 and 3. Find such pairs of
prime number upto 100.
Ans. 17 and 71; 37 and 73; 79 and 97 are such pairs of prime
numbers up to 100.
Q.4. Write down separately the prime and composite
numbers less than 20.
Ans. Prime numbers less than 20 are : 2, 3, 5, 7, 11, 13, 17 and
19.
[The numbers whose only factors are 1 and the number
itself]
Composite numbers less than 20 are : 4, 6, 8, 9, 10, 12, 14,
15, 16 and 18. [Numbers having more than two factors]
Q.5 What is the greatest prime number between 1 and 10?
Ans. The greatest prime number between 1 and 10 is 7.
Q.6 Express the following as the sum of two odd primes :
(a) 44
(b) 36
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(c) 24
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(d) 18
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Ans. (a) 44 = 13 + 31 = 3 + 41
(b) 36 = 13 + 23 = 5 + 31
(c) 24 = 11 + 13 = 5 + 19
(d) 18 = 5 + 13 = 7 + 11
Q.7. Give three pairs of prime numbers whose difference is 2.
[Remark : Two prime numbers whose difference is 2 are
called twin primes].
Ans. (i) (3, 5) (ii) (5, 7) (iii) (11, 13).
Q.8. Which of the following numbers are prime?
(a) 23
(b) 51
(c) 37
(d) 26
Ans. (a) 23 = 23 × 1
We know, the numbers whose only factors are 1 and the
number itself are called prime numbers.
Since 23 has only two factors i.e. 1 and itself so 23 is a
prime number.
(b) 51 = 1 × 51 = 3 × 17
51 has more than two factors so it is not a prime
number.
(c) 37 = 1 × 37
37 has only two factors 1 and itself so it is a prime
number.
(d) 26 = 1 × 26 = 2 × 13
Since 26 has more than two factors so it is not a prime
numbers.
Q.9. Write seven consecutive composite numbers less than
100 so that there is no prime number between them.
Ans. 90, 91, 92, 93, 94, 95 and 96.
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Q.10. Express each of the following numbers as the sum of
three odd primes :
(a) 21
(b) 31
(c) 53
(d) 61
Ans. (a) 21 = 3 + 5 + 13
(b) 31 = 3 + 5 + 23
(c) 53 = 3 + 7 + 43
(d) 61 = 3 + 5 + 53
Q.11. Write five pairs of prime numbers less than 20 whose
sum is divisible by 5.
Ans. (i) 2 + 3 = 5
(ii) 3 + 7 = 10
(iii) 2 + 13 = 15
(iv) 3 + 17 = 20
(v) 7 + 13 = 20
Q.12. Fill in the blanks :
(a) A number which has only two factors is called a
________________ .
(b) A number which has more than two factors is called
a ________________ .
(c) 1 is neither _______________ nor ________________.
(d) The smallest prime number is ________________.
(e) The smallest composite number is _______________ .
(f) The smallest even number is ________________.
Ans. (a) Prime number
(b) Composite number
(c) Prime, Composite
(d) 2
(e) 4
(f) 2
Exercise 3.3
Q.1. Using divisibility tests, determine which of the following
numbers are divisible by 2; by 3; by 4; by 5; by 6; by 8;
by 9; by 10; by 11 (say, yes or no) :
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Divisible by
Number
2
3
4
5
6
8
9
10
11
128
Yes No Yes No
No Yes No
No
No
990
…
…
…
…
…
…
…
…
…
1586
…
…
…
…
…
…
…
…
…
275
…
…
…
…
…
…
…
…
…
6686
…
…
…
…
…
…
…
…
…
639210
…
…
…
…
…
…
…
…
…
429714
…
…
…
…
…
…
…
…
…
2856
…
…
…
…
…
…
…
…
…
3060
…
…
…
…
…
…
…
…
…
4068l39
…
…
…
…
…
…
…
…
…
9
10
11
No
No
Ans.
Divisible by
Number
2
3
4
5
6
8
128
Yes No Yes No
990
Yes Yes No Yes Yes No Yes Yes Yes
1586
Yes No
No
No
No
No
No
275
No
No Yes No
No
No
No Yes
No
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No Yes No
No
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No
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6686
Yes No
No
No
No
No
639210
Yes Yes No Yes Yes No
429714
Yes Yes No
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No
No
No
No Yes Yes
No Yes No Yes No
2856
Yes Yes Yes No Yes Yes No
3060
Yes Yes Yes Yes Yes No Yes Yes No
4068l39
No Yes No
No
No
No
No
No
No
No
No
No
Q.2. Using divisibility tests, determine which of the following
numbers are divisible by 4; by 8 :
(a) 572
(b) 726352
(c) 5500
(d) 6000
(e) 12159
(f) 14560
(g) 21084
(h) 31795072
(i) 1700
(j) 2150
Ans. A number with 3 or more digits is divisible by 4, if the
number formed by its last two digits (i.e. ones and tens) is
divisible by 4.
Therefore, numbers divisible by 4 are :
572, 726352, 5500, 6000, 14560, 21084, 31795072 and
1700
A number with 4 or more digits is divisible by 8, if the
number formed by the last three digits is divisible by 8.
Therefore, numbers divisible by 8 are :
726352, 6000, 14560 and 31795072
Q.3. Using divisibility tests, determine which of the following
numbers are divisible by 6 :
(a) 297144 (b) 1258
(c) 4335
(d) 61233
(e) 901352 (f) 438750
(g) 1790184 (h) 12583
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(i) 639210 (j) 17852
Ans. If a number is divisible by 2 and 3 both then it is divisible
by 6 also.
Therefore, numbers divisible by 6 are :
297144, 438750, 1790184, 639210
Numbers not divisible by 6 are :
1258, 4335, 61233, 901352, 12583, 17852
Q.4. Using divisibility tests, determine which of the following
numbers are divisible by 11 :
(a) 5445
(b) 10824
(c) 7138965 (d) 70169308
(e) 10000001
(f) 901153
Ans. If the difference between the sum of the digits at odd places
(from the right) and the sum of the digits at even places
(from the right) of a number is either 0 or divisible by 11,
then the number is divisible by 11.
Therefore, numbers divisible by 11 are
5545, 10824, 70169308, 70169308, 10000001, 901153
Q.5. Write the smallest digit and the largest digit in the blank
space of each of the following numbers so that the
number is divisible by 3 :
(a) – 6724
(b) 4765 – 2
Ans. (a) – 6724
The given number will be divisible by 3 if the sum of
the digits of the number is divisible by 3.
Let the digit at blank space = x. Therefore, the number is
x 6724.
The sum of the digits x 6724 = x + 6 + 7 + 2 + 4 = 19 + x
The smallest value of x for which 19 + x is divisible by 3
is 2
(∵ 19 + 2 = 21, which is divisible by 3)
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∴ The required number = 26724
Similarly, the largest value of x for which 19 + x is
divisible by 3 is 8
(∵ 19 + 8 = 27, which is divisible by 3)
∴ The required number = 86724
(b) 4765 – 2
Let the digit at blank space be x. Therefore, the number
is 4765 x 2
The sum of the digits in 4765 x 2 = 4 + 7 + 6 + 5 + x + 2
= 24 + x
The smallest value of x for which 24 + x is divisible by 3
is 0
(∵ 24 + 0 = 24, which is divisible by 3)
∴ The required number = 476502
The largest value of x for which 24 + x is divisible by 3
is 9.
(∵ 24 + 9 = 33, which is divisible by 3)
∴ The required number = 476592
Q.6. Write digit in the blank space of each of the following
numbers so that the number is divisible by 11 :
(a) 92 – 389
(b) 8 – 9484
Ans. (a) 92 – 389
Let the digit at the blank space be x. Therefore, the
given number is 92 x 389
By the divisibility test of 11.
Sum of the digits at odd places = 9 + 3 + 2 = 14
Sum of the digits at even places = 8 + x + 9 = 17 + x
Difference of digits at even and odd places
= 17 + x – 14 = 3 + x
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Now, according to the divisibility test 3 + x should be
either 0 or 11.
⇒ 3 + x = 11
(3 + x = 0 given x = – 3
which is not possible)
⇒ x=8
Therefore, the required digit is 8 and the given number
is 928389.
(b) 8 – 9484
Let the digit at the blank space be x.
Therefore, the given number is 8 x 9484
By the divisibility lest of 11
Sum of the digits at odd places = 4 + 4 + x = 8 + x
Sum of the digits at even places = 8 + 9 + 8 = 25
Difference of digits at even and odd places
= (25 – (8 + x)
= 17 – x
Now, according to the divisibility lest, 17 – x should be
either 0 or 11
⇒ 17 – x = 11
(17 – x = 0, gives x = 17
which is not possible)
⇒ x=6
Exercise 3.4
Q.1. Find the common factors of :
(a) 20 and 28
(b) 15 and 25
(c) 35 and 50
(d) 56 and 120
Ans. (a) 20 = 1 × 20
28 = 1 × 28
= 2 × 10
= 2 × 14
=4×5
=4×7
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So, factors of 20 are : 1, 2, 4, 5, 10, 20.
and, factors of 28 are : 1, 2, 4, 7, 14, 28
Hence, common factors of 20 and 28 are :
1, 2 and 4. Ans.
(b) 15 = 1 × 15
25 = 1 × 25
=3×5
=5×5
So, factors of 15 are : 1, 3, 5, 15
and, factors of 25 are : 1, 5, 25
Hence, common factors of 15 and 25 are : 1 and 5 Ans.
(c) 35 = 1 × 35
50 = 1 × 50
=5×7
= 2 × 25
= 5 × 10
So, factors of 35 are : 1, 5, 7, 35
and, factors of 50 are : 1, 2, 5, 10, 25, 50
Hence, common factors of 35 and 50 are : 1 and 5. Ans.
(d) 56 = 1 × 56
= 2 × 28
= 4 × 14
=7×8
120 = 1 × 120
= 2 × 60
= 3 × 40
= 4 × 30
= 5 × 24
= 6 × 20
= 8 × 15
= 10 × 12
So, factors of 56 are : 1, 2, 4, 7, 8, 14, 28, 56
and, factors of 120 are : 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20,
24, 30, 40, 60, 120
Hence, common factors of 56 and 120 are :
1, 2, 4 and 8. Ans.
Q.2. Find the common factors of :
(a) 4, 8 and 12
(b) 5, 15 and 25
Ans. (a) Factors of 4 are : 1, 2, 4.
Factors of 8 are : 1, 2, 4, 8.
Factors of 12 are : 1, 2, 3, 4, 6, 12.
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Hence, common factors of 4, 8 and 12 are :
1, 2 and 4. Ans.
(b) Factors of 5 are : 1, 5.
Factors of 15 are : 1, 3, 5, 15
Factors of 25 are : 1, 5, 25
Hence, common factors of 5, 15 and 25 are : 1 and 5
Ans.
Q.3. Find first three common multiples of :
(a) 6 and 8
(b) 12 and 18
Ans. (a) Multiples of 6 are : 6, 12, 18, 24, 30, 36, 42, 48, 54, 60,
66, 72, …
Multiples of 8 are : 8, 16, 24, 32, 40, 48, 56, 64, 72, 80,
…
So, first three common multiples of 6 and 8 are :
24, 48 and 72
(b) Multiples of 12 are : 12, 24, 36, 48, 60, 72, 84, 96, 108,
120, …
Multiples of 18 are : 18, 36, 54, 72, 90, 108, 126, 144,
162, 180, …
So, first three common multiples of 12 and 18 are : 36,
72 and 108 Ans.
Q.4. Write all the numbers less than 100 which are common
multiples of 3 and 4.
Ans. All those numbers which are multiples of 3 × 4 = 12 will be
the common multiples of 3 and 4
Therefore, common multiple of 3 and 4 = multiple of 12
So, common multiples of 3 and 4 less than 100 are : 12, 24,
36, 48, 60, 72, 84 and 96.
Q.5. Which of the following numbers are co-prime?
(a) 18 and 35
(b) 15 and 37
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(c) 30 and 415
(d) 17 and 68
(e) 216 and 215
(f) 81 and 16
Ans. (a) Factors of 18 are : 1, 2, 3, 6, 9, 18.
Factors of 35 are : 1, 5, 7, 35.
Common factor of 18 and 35 is 1
Hence, 18 and 35 are co-primes.
(b) Factors of 15 are : 1, 3, 5, 15.
Factors of 37 are : 1, 37.
Common factor of 15 and 37 is 1.
Hence, 15 and 37 are co-prime.
(c) Factors of 30 are : 1, 2, 3, 5, 6, 15, 30.
Factors of 415 are : 1, 5, 83, 415.
Common factors of 30 and 415 is 1, 5.
Hence, 30 and 415 are not co-prime.
(d) Factors of 17 are : 1, 17.
Factors of 68 are : 1, 2, 4, 17, 34, 68.
Common factors of 17 and 68 is 1, 17.
Hence, 17 and 68 are not co-prime.
(e) Factors of 216 are : 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 27, 36,
54, 72, 216.
Factors of 215 are : 1, 5, 43, 215.
Common factor of 216 and 215 is 1.
Hence, 216 and 215 are co-prime.
(f) Factors of 81 are : 1, 3, 9, 27, 81.
Factors of 16 are : 1, 2, 4, 8, 16.
Common factor of 81 and 16 is 1.
Hence, 81 and 16 are co-prime.
Q.6. A number is divisible by both 5 and 12. By which other
number will that number always be divisible?
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Ans. ∵ 5 and 12 are co-prime.
∴ A number divisible by both 5 and 12 will always be
divisible by their multiple 5 × 12 = 60 Ans.
Q.7. A number is divisible by 12. By what other numbers will
that number be divisible?
Ans. Factors of 12 other than 12 are : 1, 2, 3, 4, 6.
So, if a number is divisible by 12, it will also be divisible by
1, 2, 3, 4 and 6. Ans.
Exercise 3.5
Q.1. Which of the following statements are true?
(a) If a number is divisible by 3, it must be divisible by
9.
(b) If a number is divisible by 9, it must be divisible by
3.
(c) A number is divisible by 18, if it is divisible by both
3 and 6.
(d) If a number is divisible by 9 and 10 both, then it
must be divisible by 90.
(e) If two numbers are co-primes, at least one of them
must be prime.
(f) All that numbers divisible by 4 must also be divisible
by 8.
(g) All that numbers divisible by 8 must also be divisible
by 4.
(h) If a number exactly divides two numbers separately,
it must exactly divide their sum.
(i) If a number exactly divides the sum of two numbers,
it must exactly divide the two numbers separately.
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Ans. (a) False
(b) True
(c) False
(d) True
(e) False
(f) False
(g) True
(h) True
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(i) False
Q.2. Here are two different factor trees for 60. Write the
missing numbers.
(a)
(b)
Ans. (a) 2 × 3 = 6
5 × 2 = 10
(b) 2 × 5 =10
10 × 3 = 30
30 × 2 = 60
Q.3. Which factors are not included in the prime
factorization of a composite number?
Ans. Let us consider any composite number say 12.
Prime factorization of 12 = 2 × 2 × 3
Therefore, in the prime factorization of a composite number
1 and the number itself are not included.
Q.4. Write the greatest four-digit number and express it in
terms of its prime factors.
Ans. Greatest four digit number = 9999
Prime factorization of 9999
= 3 × 3 × 11 × 101
Q.5. Write the smallest five-digit number and express it in
the form of its prime factors.
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Ans. The smallest five digit number = 10000
Prime factorization of 10000 = 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5
Q.6. Find all the prime factors of 1729 and arrange them in
ascending order. Now state the relation, if any; between
two consecutive prime factors.
Ans.
Prime factors of 1729 are 7, 13, 19.
The difference between two consecutive prime factors is 6.
Q.7. The product of three consecutive numbers is always
divisible by 6. Verify this statement with the help of
some examples.
Ans. (i) Let us take three consecutive numbers 5, 6, 7.
Their product = 5 × 6 × 7 = 210 which is divisible by 6.
(ii) Let us take three consecutive numbers 9, 10, 11.
Their product = 9 × 10 × 11 = 990 which is divisible by
6.
Q.8. The sum of two consecutive odd numbers is divisible by
4. Verify this statement with the help of some examples.
Ans. (i) Let us take two consecutive odd numbers 1 and 3.
Their sum 1 + 3 = 4 which is divisible by 4.
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(ii) Let us take two consecutive odd numbers 3 and 5.
Their sum 3 + 5 = 8 which is divisible by 4.
Q.9. In which of the following expressions,
factorization has been done ?
(a) 24 = 2 × 3 × 4
(b) 56 = 7 × 2 × 2 × 2
(c) 70 = 2 × 5 × 7
(d) 54 = 2 × 3 × 9
Ans. (b) 56 = 1 × 7 × 2 × 2 × 2
prime
In this expression prime factorization has been done.
(c) 70 = 2 × 5 × 7
In this expression prime factorization has been done.
Q.10. Determine if 25110 is divisible by 45.
Ans. The unit's digit of 25110 is 0.
So, 25110 is divisible by 5.
Also, sum of the digits = 2 + 5 + 1 + 1 + 0 = 9, which is
divisible by 9.
So, 25110 is divisible by 9.
Now, 5 and 9 are co-prime numbers.
Hence, 25110 is divisible by their product 5 × 9 = 45.
Q.11. 18 is divisible by both 2 and 3. It is also divisible by 2 × 3
= 6. Similarly, a number is divisible by both 4 and 6.
Can we say that the number must also be divisible by 4
× 6 = 24? If not, give an example to justify your answer.
Ans. No, because 4 and 6 are not co-primes.
e.g. (i) 36 is divisible by both 4 and 6. But it is not
divisible by 24.
(ii) 12 is divisible by both 4 and 6 but 12 is not
divisible by 24.
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Q.12. I am the smallest number having four different prime
factors. Can you find me?
Ans. The smallest four different prime factors are 2, 3, 5, 7.
Hence, the smallest number, having four different prime
factors = 2 × 3 × 5 × 7 = 210.
Exercise 3.6
Q.1. Find the H.C.F. of the following numbers :
(a) 18, 48
(b) 30, 42
(c) 18, 60
(d) 27, 63
(e) 36, 84
(f) 34, 102
(g) 70, 105, 175
(h) 91, 112
(i) 18, 54, 81
(j) 12, 45, 75
Ans. (a) Factors of 18 are : 2 × 3 × 3
Factors of 48 are : 2 × 2 × 2 × 2 × 3
HCF of 18 and 48 is 2 × 3 = 6
(b) Factors of 30 are : 2 × 3 × 5
Factors of 42 are : 2 × 3 × 7
HCF 2 × 3 = 6
(c) Factors of 18 are : 2 × 3 × 3
Factors of 60 are : 2 × 2 × 3 × 5
HCF 2 × 3 = 6
(d) Factors of 27 are : 3 × 3 × 3
Factors of 63 are : 3 × 3 × 7
HCF 3 × 3 = 9
(e) Factors of 36 are : 2 × 2 × 3 × 3
Factors of 84 are : 2 × 2 × 3 × 7
HCF 2 × 2 × 3 = 12
(f) Factors of 34 are : 2 × 17 × 1
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Q.2.
Ans.
Q.3.
Ans.
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Factors of 102 are : 2 × 3 × 17
HCF 2 × 17 = 34
(g) Factors of 70 are : 2 × 5 × 7
Factors of 105 are : 3 × 5 × 7
Factors of 175 are : 5 × 5 × 7
HCF 5 × 7 = 35
(h) Factors of 91 are : 7 × 13
Factors of 112 are : 2 × 2 × 2 × 2 × 7
HCF = 7
(i) Factors of 18 are : 2 × 3 × 3
Factors of 54 are : 2 × 3 × 3 × 3
Factors of 81 are : 3 × 3 × 3 × 3
HCF 3 × 3 = 9
(j) Factors of 12 are : 2 × 2 × 3
Factors of 45 are : 3 × 3 × 5
Factors of 75 are : 3 × 5 × 5
HCF = 3
What is the H.C.F. of two consecutive :
(a) numbers?
(b) even numbers?
(c) odd numbers?
(a) H.C.F. of two consecutive numbers is 1.
(b) H.C.F. of two consecutive even numbers is 2.
(c) H.C.F. of two consecutive odd numbers is 1.
H.C.F. of co-prime numbers 4 and 15 was found as
follows by factorization : 4 = 2 × 2 and 15 = 3 × 5, since
there is no common factor, so HCF of 4 and 15 is 0. Is
the answer correct? If not, what is the correct HCF?
No, the correct HCF is 1.
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Exercise 3.7
Q.1. Renu purchases two bags of fertiliser of weights 75 kg
and 69 kg. Find the maximum value of weight which can
measure the weight of the fertiliser exact number of
times.
Ans. Factor of 75 are 3 × 5 × 5
Factor of 69 are 3 × 23
HCF = 3
Hence, maximum value of weight is 3 kg.
Q.2. Three boys step off together from the same spot. Their
steps measure 63 cm, 70 cm and 77 cm respectively.
What is the minimum distance each should cover so that
all can cover the distance in complete steps?
Ans. Factor of 63 cm are 3 × 3 × 7
Factor of 70 cm are 2 × 5 × 7
Factors of 77 cm are 7 × 11
LCM of 63, 70 and 77 is
= 7 × 2 × 5 × 9 × 11
= 70 × 99
= 6930 cm
Hence, required minimum distance is 6930 cm. Ans.
Q.3. The length, breadth and height of a room are 825 cm,
675 cm and 450 cm respectively. Find the longest tape
which can measure the three dimensions of the room
exactly.
Ans. Factor of 825 are 5 × 5 × 3 × 11
Factors of 450 are 3 × 3 × 2 × 5 × 5
Factors of 675 are 5 × 5 × 3 × 3 × 3
HCF = 5 × 5 × 3 = 75
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Q.4.
Ans.
Q.5.
Ans.
Q.6.
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Hence required longest tape will be 75 cm.
Determine the smallest 3-digit number which is exactly
divisible by 6, 8 and 12.
Factors of 6 = 2 × 3
Factors of 8 = 2 × 2 × 2
Factors of 12 = 2 × 2 × 3
LCM = 2 × 2 × 2 × 3 = 24
Smallest 3-digit number = 100
On dividing 100 by 24 we get,
4 as remainder
Hence the required number = 100 + 24 – 4 = 120. Ans.
Determine the largest 3-digit number exactly divisible
by 8, 10 and 18.
Factors of 8 = 2 × 2 × 2
Factors of 10 = 2 × 5
Factors of 12 = 2 × 2 × 3
LCM = 2 × 2 × 2 × 3 × 5
= 120
Now largest 3 digit number = 999
Hence, required number = 999 – 39 = 960 Ans.
The traffic lights at three different road crossing change
after every 48 seconds, 72 seconds and 108 seconds
respectively. If they change simultaneously at 7 A.M., at
what time they will change simultaneously again?
Factors of 48 sec are : 2 × 2 × 2 × 2 × 3
Factors of 72 sec are : 2 × 2 × 2 × 3 × 3
Factors of 108 sec are : 2 × 2 × 3 × 3 × 3
LCM = 2 × 2 × 2 × 2 × 3 × 3 × 3
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= 432 sec.
Q.7.
Sol.
Q.8.
Ans.
Q.9.
Ans.
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(∵ 1 minute = 60 )
= 7 min 12 sec. Ans.
Three tankers contain 403 litres, 434 litres and 465 litres
of diesel respectively. Find the maximum capacity of a
container that can measure the diesel of three containers
exact number of times.
The maximum capacity of a container will be the HCF of
403, 434, 465.
Factors of 403 = 31 × 13
Factors of 465 = 31 × 5 × 3
Factors of 434 = 31 × 7 × 2
HCF = 31
Hence, maximum capacity of the required container
= 31 litres.
Find the least number which when divided by 6, 15 and
18 leave remainder 5 in each case.
Factor of 6 = 2 × 3
Factor of 15 = 3 × 5
Factor of 18 = 2 × 3 × 3
LCM = 2 × 3 × 3 × 5 = 90
Hence, the required number = 90 + 5 = 95 Ans.
Find the smallest four-digit number which is divisible by
18, 24 and 32.
Factor of 18 are 2 × 3 × 3
Factor of 24 are 2 × 2 × 2 × 3
Factor of 32 are 2 × 2 × 2 × 2 × 2
LCM = 2 × 2 × 2 × 2 × 3 × 3 × 2
= 288
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The smallest 4 digit number = 1000
On dividing 1000 by 288 we get, 136 as remainder
Hence the required number = 1000 + 288 – 136 = 1152
Q10. Find the LCM of the following numbers
(a) 9 and 4
(b) 12 and 5
(c) 6 and 5
(d) 15 and 4
Observe a common property in the obtained LCM Is
LCM the product of two numbers in each case?
Ans. (a) Factors of 4 are 2 × 2
Factors of 9 are 3 × 3
LCM of 4 and 9 = 2 × 2 × 3 × 3 = 36
(b) Factors of 12 are 2 × 2 × 3
Factors of 5 are 1 × 5
LCM of 12 and 5 = 1 × 2 × 2 × 3 × 5
= 60
(c) Factors of 6 and 5 are
Factors of 6 are 2 × 3
Factors of 5 are 1 × 5
LCM of 6 and 5 = 2 × 3 × 5 = 30
(d) Factors of 15 are 3 × 5
Factors of 4 are 2 × 2
LCM of 15 and 4 = 3 × 5 × 2 × 2
= 60
Yes in each case LCM is the product of two numbers
Q11. Find the LCM of the following numbers in which one
number is the factor of the other
(a) 5, 20
(b) 6, 18
(c) 12, 48
(d) 9, 45
What do you observe in the results obtained?
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Ans.
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(a) LCM of 5 and 20
Factors of 5 are 5 × 1
Factors of 20 are 5 × 4
LCM = 4 × 5 = 20
(b) Factors of 6 are 2 × 3
Factors of 18 are 2 × 3 × 3
LCM = 2 × 3 × 3 = 18
(c) Factors of 12 are 2 × 2 × 3
Factors of 48 are 2 × 2 × 2 × 2 × 3
LCM = 2 × 2 × 3 × 2 × 2
= 16 × 3 = 48
(d) Factors of 9 are 3 × 3
Factors of 45 are 3 × 3 × 5
LCM = 9 × 5 = 45
We observe that greater number is the LCM in each
case.
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