Exercise 3.1
1.
Write all the factors
of the following numbers:
The factor of a number is an
exact divisor of that number.
a)
24
Solution
1 × 24 = 24 2 × 12 = 24 3 × 8 = 24 4 × 6 =
24
Thus, the factors of 24 are: 1,
2, 3, 4, 6, 8, 12, 24
b)
15
Solution
1 × 15 = 15 3 × 5 = 15
Factors of 15 are 1, 3, 5 and 15
c)
21
Solution
1 × 21= 21 3 × 7 = 21
Factors of 21 are 1, 3, 7 and 21.
d)
27
Solution
1 × 27 = 27 3 × 9 = 27
Factors of 27 are 1, 3, 9 and 27
e)
12
Solution
1 × 12 = 12 2 × 6 = 12 3
× 4 = 12
Factors of 12 are 1, 2, 3, 4, 6
and 12
f)
20
1 × 20 = 20 2 × 10 = 20 4 × 5 = 20
Factors of 20 are 1, 2, 4, 5, 10,
20
g)
18
1 × 18 = 18 2 × 9 = 18 3
× 6 = 18
Factors of 18 are 1, 2, 3, 6, 9,
18
h)
23
1 × 23 = 23
Factors of 23 are 1 and 23. [23 is
a prime number]
i)
36
1 × 36 = 36 2 × 18 = 36 3 × 12 = 36 4 × 9 =
36 6 × 6 = 36
Factors of 36 are 1, 2, 3, 4, 6,
9, 12, 18 and 36
2.
Write first five
multiples of :
a)
5
Solution
5 × 1 = 5 5 × 2 = 10 5 ×
3 = 15 5 × 4 =20 5 × 5= 25
The first five multiples of 5 are
5, 10, 15, 20 and 25
b)
8
8 × 1 = 8 8 × 2 = 16 8 ×
3 = 24 8 × 4 = 32 8 × 5= 40
The first five multiples of 8 are
8, 16, 24, 32 and 40
c)
9
9 × 1 = 9 9 × 2 = 10 9 ×
3 = 27 9 × 4 = 36 9 × 5= 45
The first five multiples of 9 are
9, 18, 27, 36 and 45
3.
Match the items in
column 1 with the items in column 2.
Column 1
|
Column 2
|
(i) 35
|
(a) Multiple of 8
|
(ii) 15
|
(b) Multiple of 7
|
(iii) 16
|
(c) Multiple of 70
|
(iv) 20
|
(d) Factor of 30
|
(v) 25
|
(e) Factor of 50
|
(f) Factor of 20
|
Solution
i.
35 is a multiple of 7. (b)
ii.
15 is a factor of 30 (d)
iii.
16 is a multiple of 8 (a)
iv.
20 is a factor of 20 (f)
v.
25 is a factor of 50 (e)
4.
Find all the multiples of 9 up to 100.
Solution
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 9 × 12 = 108
Multiples of 9 up
to 100 are: 9, 18, 27, 36, 45, 54, 63, 72, 81, 90 and 99
Exercise 3.2
1.
What is the sum of
any two (a) Odd numbers? (b) Even numbers?
Solution
Consider any two odd numbers say
7 and 13.
7 + 13 = 20.
Similarly, 23 + 21 = 44
20 and 44 are both even numbers.
Thus, the sum of any two odd
numbers is an even number.
Now, consider even numbers 12 and
14. 12 + 14 = 28.
Similarly, 24 + 64 = 88
28 and 88 are even numbers.
Thus, the sum of any two even
numbers is an even number.
2.
State whether the
following statements are True or False:
(a)
The sum of three
odd numbers is even.
False. [Ex: 7 + 3 + 9 = 19.]
(b)
The sum of two odd
numbers and one even number is even.
True. [ex: 7 + 3 + 4 = 14.]
(c)
The product of
three odd numbers is odd.
True. [Ex: 7 × 3 × 9 = 21 × 9 =
189]
(d)
If an even number
is divided by 2, the quotient is always odd.
False. [Ex: 32 ÷ 4 = 8]
(e)
All prime numbers
are odd.
False, every prime number except
2 is odd.
(f)
Prime numbers do
not have any factors.
False, prime numbers have two
factors, the number 1 and itself.
(g)
Sum of two prime
numbers is always even.
False. [3 + 2 = 5]
(h)
2 is the only even
prime number.
True
(i)
All even numbers
are composite numbers.
False, 2 is an even prime number.
(j)
The product of two
even numbers is always even.
True. [Ex: 12 × 4 = 36]
3.
The numbers 13 and
31 are prime numbers. Both these numbers have same digits 1 and 3. Find such
pairs of prime numbers up to 100.
Solution
17 and 71
37 and 73
79 and 97
4.
Write down
separately the prime and composite numbers less than 20.
Solution
Prime numbers less than 20:
2, 3, 5, 7, 11, 13, 17, 19
Composite numbers less than 20:
4, 6, 8, 9, 10, 12, 14, 15, 16,
18
5.
What is the
greatest prime number between 1 and 10?
Solution
Prime numbers between 1 and 10
are: 2, 3, 5, 7.
Greatest prime number between 1
and 10 is 7.
6.
Express the
following as the sum of two odd primes.
[There are many ways of writing a
number as the sum of two odd primes. The following solutions show a few of
them]
(a)
44
Solution
44 = 41 + 3 OR 37 + 7
(b)
36
Solution
36 = 31 + 5 OR 29 + 7
(c)
24
Solution
24 = 17 + 7 OR 19 + 5
(d)
18
Solution
18 = 11 + 7 OR 13 + 5
7.
Give three pairs of
prime numbers whose difference is 2. [Remark: Two prime numbers whose
difference is 2 are called twin primes].
Solution
3 and 5
5 and 7
11and 13
8.
Which of the
following numbers are prime?
a)
23
b)
51
c)
37
d)
26
Solution
The factors of 23
are 1 and 23.
The factors of 51
are 1, 51, 13 and 7.
The factors of 37
are 1 and 37
The factors of 26
are 1, 26, 13, 2
Prime numbers are
numbers whose only factors are 1 and itself.
So, a) and c) are
prime numbers.
9.
Write seven
consecutive composite numbers less than 100 so that there is no prime number
between them.
Solution
90, 91, 92, 93, 94, 95, 96
10. Express each of the following
numbers as the sum of three odd primes:
(a)
21
Solution
21 = 7 + 9 + 5
(b)
31
Solution
31 = 23 + 5 + 3
(c)
53
Solution
53 = 23 + 17 + 13
(d)
61
Solution
61 = 47 + 11 + 3
11. Write five pairs of prime numbers
less than 20 whose sum is divisible by 5. (Hint : 3+7 = 10)
Solution
Prime numbers less than 20 are 2,
3, 5, 7, 11, 13, 17, 19.
Let us write five pairs of prime
numbers whose sum is divisible by 5.
2 + 3 = 5
2 + 13 = 15
3 + 17 = 20
7 + 13 = 20
11 + 19 = 30
12. Fill in the blanks :
(a)
A number which has
only two factors is called a ______.
Prime number.
(b)
A number which has
more than two factors is called a ______.
Composite number
(c)
1 is neither ______
nor ______.
Prime, composite
(d)
The smallest prime
number is ______.
2
(e)
The smallest
composite number is _____.
4
(f)
The smallest even number is ______.
2
Exercise 3.3
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):
Solution
990
Divisible by 2 as it has 0 in its
ones place.
Divisible by 3 as the sum of the
digits, 18, is a multiple of 3.
Not divisible by 4 as the number
formed by its last two digits, 90, is not divisible by 4.
Divisible by 5 as the number has
0 in its ones place.
Divisible by 6 as the number is
divisible by both 2 and 3.
Not divisible by 8.
Divisible by 9 as the sum of the
digits of the number, 9 + 9 = 18, is divisible by 9.
Divisible by 10 as the number has
0 in its units place.
The number is divisible by 11.
The sum of the digits in the odd
places from the right is 9 + 0 = 9. The sum of the digits in the even places from
the right is 9. The difference is 0. If the difference is 0 or a divisible by
11, then the number is divisible by 11.
1586
Divisible by 2 as the number in
the units place is 2.
Not divisible by 3 as the sum of
the digits, 20, is not a multiple of 3.
Not divisible by 4 as the number
formed by its last two digits, 86, is not divisible by 4.
Not divisible by 5 as the number does
not have 0 or 5 in its ones place.
Not divisible by 6 as the number
is divisible by 2 but not by 3.
Not divisible by 8 as the number
formed by the last three digits, 586, is not divisible by 8.
Not divisible by 9 as the sum of
the digits of the number, 20, is not divisible by 9.
Not divisible by 10 as the number
does not have 0 in its units place.
The number is not divisible by
11.
The sum of the digits in the odd
places from the right is 6 + 5 = 11. The sum of the digits in the even places from
the right is 8 + 1 = 9. The difference is 2. If the difference is 0 or a
divisible by 11, then the number is divisible by 11. So, the number is not
divisible by 11.
275
Not divisible by 2 as the number does
not have 0, 2, 4, 6 or 8 in its units place.
Not divisible by 3 as the sum of
the digits, 14, is not a multiple of 3.
Not divisible by 4 as the number
formed by its last two digits, 75, is not divisible by 4.
Divisible by 5 as the number has 5
in its ones place.
Not divisible by 6 as the number
is neither divisible by 2 nor by 3.
Not divisible by 8.
Not divisible by 9 as the sum of
the digits of the number, 14, is not divisible by 9.
Not divisible by 10 as the number
does not have 0 in its units place.
The number is divisible by 11.
The sum of the digits in the odd
places from the right is 5 + 2 = 7. The sum of the digits in the even places
from the right is 7. The difference is 0. If the difference is 0 or a divisible
by 11, then the number is divisible by 11. So, the number is divisible by 11.
6686
Divisible by 2 as the number has
6 in its units place.
Not divisible by 3 as the sum of
the digits, 26, is not a multiple of 3.
Not divisible by 4 as the number
formed by its last two digits, 86, is not divisible by 4.
Not divisible by 5 as the number does
not have 0 or 5 in its ones place.
Not divisible by 6 as the number
is divisible by 2 but not by 3.
Not divisible by 8 as the number
formed by the last three digits 686 is not divisible by 8.
Not divisible by 9 as the sum of
the digits of the number, 86, is not divisible by 9.
Not divisible by 10 as the number
does not have 0 in its units place.
The number is not divisible by
11.
The sum of the digits in the odd
places from the right is 6 + 6 = 12. The sum of the digits in the even places
from the right is 14. The difference is 2. If the difference is 0 or a number divisible
by 11, then the number is divisible by 11. So, the number is not divisible by
11.
639210
Divisible by 2 as the number has 0
in its units place.
Divisible by 3 as the sum of the
digits, 21, is a multiple of 3.
Not divisible by 4 as the number
formed by its last two digits, 10, is not divisible by 4.
Divisible by 5 as the number has 0
in its ones place.
Divisible by 6 as the number is
divisible by 2 and by 3.
Not divisible by 8 as the number
formed by the last three digits, 210, is not divisible by 8.
Not divisible by 9 as the sum of
the digits of the number, 21, is not divisible by 9.
Divisible by 10 as the number has
0 in its units place.
The number is divisible by 11.
The sum of the digits in the odd
places from the right is 5. The sum of the digits in the even places from the
right is 16. The difference is 11. If the difference is 0 or a divisible by 11,
then the number is divisible by 11. So, the number is divisible by 11.
429714
Divisible by 2 as the number has
4 in its units place.
Divisible by 3 as the sum of the
digits, 27, is a multiple of 3.
Not divisible by 4 as the number
formed by its last two digits, 14, is not divisible by 4.
Not divisible by 5 as the number does
not have 0 or 5 in its ones place.
Divisible by 6 as the number is
divisible by both 2 and by 3.
Not divisible by 8 as the number
formed by the last three digits, 714, is not divisible by 8.
Divisible by 9 as the sum of the
digits of the number, 27, is divisible by 9.
Not divisible by 10 as the number
does not have 0 in its units place.
The number is not divisible by
11.
The sum of the digits in the odd
places from the right is 13. The sum of the digits in the even places from the
right is 14. The difference is 1. If the difference is 0 or a divisible by 11,
then the number is divisible by 11. So, the number is not divisible by 11.
2856
Divisible by 2 as the number has 6
in its units place.
Divisible by 3 as the sum of the
digits, 21, is a multiple of 3.
Divisible by 4 as the number
formed by its last two digits, 56, is divisible by 4.
Not divisible by 5 as the number does
not have 0 or 5 in its ones place.
Divisible by 6 as the number is
divisible by both 2 and by 3.
Divisible by 8 as the number
formed by the last three digits, 856, is divisible by 8.
Not divisible by 9 as the sum of
the digits of the number, 21, is not divisible by 9.
Not divisible by 10 as the number
does not have 0 in its units place.
The number is not divisible by
11.
The sum of the digits in the odd
places from the right is 14. The sum of the digits in the even places from the
right is 7. The difference is 7. If the difference is 0 or a divisible by 11,
then the number is divisible by 11. So, the number is not divisible by 11.
3060
Divisible by 2 as the number has 0
in its units place.
Divisible by 3 as the sum of the
digits, 9, is a multiple of 3.
Divisible by 4 as the number
formed by its last two digits, 60, is divisible by 4.
Divisible by 5 as the number has 0
in its ones place.
Divisible by 6 as the number is
divisible by both 2 and by 3.
Not divisible by 8 as the number
formed by the last three digits, 60, is not divisible by 8.
Divisible by 9 as the sum of the
digits of the number, 9, is divisible by 9.
Divisible by 10 as the number has
0 in its units place.
The number is not divisible by
11.
The sum of the digits in the odd
places from the right is 0. The sum of the digits in the even places from the
right is 9. The difference is 9. If the difference is 0 or a divisible by 11,
then the number is divisible by 11. So, the number is not divisible by 11.
406839
Not divisible by 2 as the number
does not have 0, 2, 4, 6 or 8 in its units place.
Divisible by 3 as the sum of the
digits, 30, is a multiple of 3.
Not divisible by 4 as the number
formed by its last two digits, 39, is not divisible by 4.
Not divisible by 5 as the number does
not have 0 or 5 in its ones place.
Not divisible by 6 as the number
is not divisible by 2.
Not divisible by 8 as the number
formed by the last three digits, 839, is not divisible by 8.
Not divisible by 9 as the sum of
the digits of the number, 30, is not divisible by 9.
Not divisible by 10 as the number
does not have 0 in its units place.
The number is not divisible by
11.
The sum of the digits in the odd
places from the right is 17. The sum of the digits in the even places from the
right is 13. The difference is 4. If the difference is 0 or a divisible by 11,
then the number is divisible by 11. So, the number is not divisible by 11.
Number
|
Divisible by
|
||||||||
2
|
3
|
4
|
5
|
6
|
8
|
9
|
10
|
11
|
|
128
|
yes
|
no
|
yes
|
No
|
no
|
yes
|
no
|
no
|
no
|
990
|
yes
|
yes
|
no
|
Yes
|
yes
|
no
|
yes
|
yes
|
Yes
|
1586
|
yes
|
no
|
no
|
No
|
no
|
no
|
no
|
no
|
No
|
275
|
No
|
no
|
no
|
Yes
|
no
|
no
|
no
|
no
|
Yes
|
6686
|
Yes
|
No
|
No
|
No
|
No
|
No
|
No
|
No
|
No
|
639210
|
Yes
|
Yes
|
No
|
Yes
|
Yes
|
No
|
No
|
Yes
|
Yes
|
429714
|
Yes
|
Yes
|
No
|
No
|
Yes
|
No
|
yes
|
No
|
No
|
2856
|
Yes
|
Yes
|
Yes
|
No
|
Yes
|
Yes
|
No
|
No
|
No
|
3060
|
Yes
|
Yes
|
Yes
|
Yes
|
Yes
|
No
|
Yes
|
Yes
|
No
|
406839
|
No
|
Yes
|
No
|
No
|
No
|
No
|
No
|
No
|
No
|
2.
Using divisibility
tests, determine which of the following numbers are divisible by 4; by 8:
A number with 3 or more digits is
divisible by 4 if the number formed by its last two digits is divisible by 4.
A number with 3 or more digits is
divisible by 8 if the number formed by its last three digits is divisible by 8.
a)
572
72 is divisible by 4, hence the
number is divisible by 4.
The number is not divisible by 8.
b)
726352
52 is divisible by 4, hence the
number is divisible by 4.
352 is divisible by 8, hence the
number is divisible by 8.
c)
5500
0 is divisible by 4, hence the
number is divisible by 4.
500 is not divisible by 8, hence
the number is not divisible by 8.
d)
6000
0 is divisible by 4 and 8. Hence,
the number is divisible by 4 and 8.
e)
12159
59 is not divisible by 4. Hence,
the number is not divisible by 4.
159 is not divisible by 8, hence
the number is not divisible by 8.
f)
14560
60 is divisible by 4, hence the
number is divisible by 4.
560 is divisible by 8, hence the
number is divisible by 8.
g)
21084
84 is divisible by 4, hence the
number is divisible by 4.
84 is not divisible by 8, hence
the number is not divisible by 8.
h)
31795072
72 is divisible by 4 and 8. Hence
the number is divisible by 4 and 8.
i)
1700
The number is divisible by 4.
700 is not divisible by 8, hence
the number is not divisible by 8.
j)
2150
50 is not divisible by 4, hence
the number is not divisible by 4.
150 is not divisible by 8, hence
the number is not divisible by 8.
3.
Using divisibility
tests, determine which of following numbers are divisible by 6:
A number is divisible by 6 if it
is divisible by both 2 and 3.
a)
297144
The number is divisible by 2, as
the number has 4 in its ones place.
The number is divisible by 3 as
the sum of the digits, 27, is divisible by 3.
Hence the number is divisible by
6.
b)
1258
The number is not divisible by 3
as the sum of the digits, 16, is not divisible by 3. Hence, the number is not
divisible by 6.
c)
4335
The number is not divisible by 2
as the number does not have 0, 2, 4, 6, or 8 in its units place. Hence the
number is not divisible by 6.
d)
61233
The number is not divisible by 2
as the number does not have 0, 2, 4, 6, or 8 in its units place. Hence the
number is not divisible by 6.
e)
901352
The number is not divisible by 3
as the sum of the digits, 20, is not divisible by 3. Hence, the number is not
divisible by 6.
f)
438750
The number is divisible by 2, as
the number has 0 in its ones place.
The number is divisible by 3 as
the sum of the digits, 27, is divisible by 3.
Hence the number is divisible by
6.
g)
1790184
The number is divisible by 2, as the
number has 4 in its ones place.
The number is divisible by 3 as
the sum of the digits, 30, is divisible by 3.
Hence the number is divisible by
6.
h)
12583
The number is not divisible by 2
as the number does not have 0, 2, 4, 6, or 8 in its units place. Hence the
number is not divisible by 6.
i)
639210
The number is divisible by 2, as the
number has 4 in its ones place.
The number is divisible by 3 as
the sum of the digits, 21, is divisible by 3.
Hence the number is divisible by
6.
j)
17852
The number is not divisible by 3
as the sum of the digits, 23, is not divisible by 3. Hence, the number is not
divisible by 6.
4.
Using divisibility
tests, determine which of the following numbers are divisible by 11:
Find 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 the number. If the difference is either 0 or
divisible by 11, then the number is divisible by 11.
a)
5445
The sum of the digits in the odd
places from the right is 9. The sum of the digits in the even places from the
right is 9. The difference is 0. The number is not divisible by 11.
b)
10824
Sum of the digits in the odd
places from the right is 13.
Sum of the digits in the even
places from the right is 2.
Difference is 11.
The number is divisible by 11.
c)
7138965
Sum of the digits in the odd
places from the right is 24.
Sum of the digits in the even
places from the right is 15.
Difference is 9.
The number is not divisible by 11.
d)
70169308
Sum of the digits in the odd
places from the right is 17.
Sum of the digits in the even
places from the right is 17.
Difference is 0.
The number is divisible by 11.
e)
10000001
Sum of the digits in the odd
places from the right is 1.
Sum of the digits in the even
places from the right is 1.
Difference is 0.
The number is divisible by 11.
f)
901153
Sum of the digits in the odd
places from the right is 4.
Sum of the digits in the even
places from the right is 15.
Difference is 11.
The number is divisible by 11.
5.
Write the smallest
digit and the greatest digit in the blank space of each of the following numbers
so that the number formed is divisible by 3:
a)
__ 6724
Solution
A number is divisible by 3, if
the sum of the digits is a multiple of 3.
The sum of the digits of the
given number is 6 + 7 + 2 + 4 + __ = 19.
The greatest digit that could be
added is 9.
But, 19 + 9 = 28, which is not a
multiple of 3.
So, find 19 + 8 = 27, which is a
multiple of 3.
The greatest digit in the
blank space should be 8.
Similarly, the smallest digit
which can be added to 19 is 0. But 19 is not a multiple of 3.
19 + 1 = 20, not a multiple of 3.
19 + 2 = 21, is a multiple of 3.
Hence, the smallest digit that
should be in the blank space is 2.
b)
4765 __ 2
Solution
The Sum of the digits is 4 + 7 +
6 + 5 + 2 = 24
The greatest digit that could be
added is 9.
24 + 9 = 33; which is a multiple of
3.
The smallest digit which can be
added to 24 is 0, as 24 is a multiple of 3.
The greatest digit to be added is
9 and the smallest digit to be added is 0.
6.
Write a digit in
the blank space of each of the following numbers so that the number formed is
divisible by 11 :
(a)
92 __ 389
Solution
Let p be the
missing number in the blank space.
Find the sum of the
digits at the odd places from the right.
9 + 3 + 2 = 14
Find the sum of the
digits at the even places from the right.
8 + p + 9 = 17 + p
If the difference
is either 0 or 11, then the number is divisible by 11.
That is, 17 + p –
14 = 0 OR 17 + p – 14 = 11
3 + p = 0 OR 3 + p
= 11
p = -3 OR p = 8
But, p cannot be
negative.
Hence, p = 8.
The missing digit
in the blank space is 8.
(b)
8 __ 9484
Solution
Let p be the
missing number in the blank space.
Find the sum of the
digits at the odd places from the right.
4 + 4 + p = 8 + p
Find the sum of the
digits at the even places from the right.
8 + 9 + 8 = 25
If the difference
is either 0 or 11, then the number is divisible by 11.
That is, 25 – (8 +
p) = 0 OR 25 – (8 + p) = 11
17 – p = 0 OR 17 –
p = 11
p = 17 OR p = 6
But, p = 17 is not
possible.
Hence, p = 6.
The missing digit
in the blank space is 6.
Exercise 3.4
Exercise 3.4
1.
Find the common
factors of:
a)
20 and 28
Solution
Factors of 20 are 1, 2, 4, 5, 10,
20
Factors of 28 are 1, 2, 4, 7, 14,
28
Common factors are: 1, 2, 4
b)
15 and 25
Solution
Factors of 15 are 1, 3, 5, 15
Factors of 25 are 1, 5, 25
Common factors are 1, 5
c)
35 and 50
Factors of 35 are 1, 5, 7, 35
Factors of 50 are 1, 2, 5, 10, 25,
50
Common factors are 1, 5,
d)
56 and 120
Factors of 56 are 1, 2, 4, 7, 8,
14, 28, 56
Factors of 120 are 1, 2, 3, 4, 5,
6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120
Common factors are 1, 2, 4, 8
2.
Find the common
factors of :
a)
4, 8 and 12
Solution
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
Common factors are 1, 2, 4
b)
5, 15 and 25
Solution
Factors of 5 are 1, 5
Factors of 15 are 1, 3, 5, 15
Factors of 25 are 1, 5, 25
Common factors are 1, 5
3.
Find first three
common multiples of :
a)
6 and 8
Solution
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, 88, 96…
First three common multiples are
24, 48, 72
b)
12 and 18
Multiples of 12 are 12, 24, 36,
48, 60, 72, 84, 108, 120, 132, 144,…
Multiples of 18 are 18, 36, 54,
72, 90, 108, 126, 144, 162, 180, 198…
First three common multiples are
36, 72, 108
4.
Write all the
numbers less than 100 which are common multiples of 3 and 4.
Solution
Multiples of 3 less than 100:
3, 6, 9, 12, 15, 18, 21, 24,…93,
96, 99
Multiples of 4 less than 100:
4, 8, 12, 16, …96
Common multiples are: 12, 24, 36,
48, 60, 72, 84, 96
5.
Which of the
following numbers are co-prime?
Two numbers having only 1 as a
common factor are called co-prime numbers.
a)
18 and 35
Solution
Factors of 18 are 1, 2, 3, 6, 9,
18
Factors of 35 are 1, 5, 7, 35
Common factor is 1.
So, 18 and 35 are Co-prime
numbers
b)
15 and 37
Solution
Factors of 15 are 1, 3, 5, 15
Factors of 37 are 1, 37
Common factor is 1.
15 and 37 are Co-prime numbers.
c)
30 and 415
Solution
Factors of 30 are 1, 2, 3, 5, 6,
10, 15, 30
Factors of 415 are 1, 5, 83, 415
Common factors are 1 and 5.
The numbers 30 and 415 are not
Co-prime
d)
17 and 68
Factors of 17 are 1, 17
Factors of 68 are 1, 2, 4, 17,
34, 68
Common factors are 1, 17
The numbers are not co-prime.
e)
216 and 215
Factors of 216 are 1, 2, 3, 4, 6,
8, 9, 12, 18, 24, 27, 36, 54, 72, 108, 216,
Factors of 215 are 1, 5, 43, 215
Common factor is 1.
The numbers are co-prime.
f)
81 and 16
Factors of 81 are 1, 3, 9, 27, 81
Factors of 16 are 1, 2, 4, 8, 16
Common factor 1
The numbers are co-prime.
6.
A number is
divisible by both 5 and 12. By which other number will that number be always
divisible?
Solution
The number is divisible by 5 and
12 and as 5 and 12 are co-prime numbers, the number must be divisible by the
product 5 × 12 = 60.
So, the given number will always
be divisible by 60.
7.
A number is
divisible by 12. By what other numbers will that number be divisible?
Solution
The number is divisible by 12.
So, the number will be divisible
by all the factors of 12.
Factors of 12 are 1, 2, 3, 4, 6,
12
The number will also be divisible
by the numbers 2, 3, 4 and 6.
Exercise 3.5
1.
Which of the
following statements are true?
a)
If a number is
divisible by 3, it must be divisible by 9.
False, 12 is divisible by 3 but
not by 9.
b)
If a number is
divisible by 9, it must be divisible by 3.
True
c)
A number is
divisible by 18, if it is divisible by both 3 and 6.
False, 12 is divisible by both 3
and 6 but not by 18.
d)
If a number is
divisible by 9 and 10 both, then it must be divisible by 90.
True
e)
If two numbers are
co-primes, at least one of them must be prime.
False, 14 and 27 are co-primes
and are composite numbers.
f)
All numbers which
are divisible by 4 must also be divisible by 8.
False, 12 is divisible by 4 but
not by 8.
g)
All numbers which
are divisible by 8 must also be divisible by 4.
True
h)
If a number exactly
divides two numbers separately, it must exactly divide their sum.
True
i)
If a number exactly
divides the sum of two numbers, it must exactly divide the two numbers
separately.
False, 3 divides 18 but not 16
and 2.
2. Here are two different factor trees for 60. Write the missing numbers.
a)
Solution
6
× 10 = 60
Now,
2 × 3 = 6 and 5 × 2 = 10
So,
the missing number are 3 and 2.
b)
Solution
60 = 30 × 2
30 = 10 × 3
10 = 2 × 5
The missing numbers are 2, 3 and 5
3.
Which factors are
not included in the prime factorisation of a composite number?
Solution
1 and the number itself.
4.
Write the greatest
4-digit number and express it in terms of its prime factors.
Solution
The greatest 4-digit number is
9999.
So, the prime factors are 3 × 3 × 11 × 101
5.
Write the smallest
5-digit number and express it in the form of its prime factors.
Solution
The smallest 5-digit number is
10,000.
The prime factors of 10,000 is 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5
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.
Solution
1729 = 7 × 13 × 19
The difference between the prime
factors is 6.
7.
The product of
three consecutive numbers is always divisible by 6. Verify this statement with
the help of some examples.
Solution
12 × 13 × 14 = 2184; is divisible
by 6.
25 × 26 × 27 = 17550; is
divisible by 6
8.
The sum of two
consecutive odd numbers is divisible by 4. Verify this statement with the help
of some examples.
Solution
3 + 5 = 8, is divisible by 4.
13 + 15 = 28, is divisible by 4.
9.
In which of the
following expressions, prime factorisation has been done?
(a)
24 = 2 × 3 × 4
4 is not a prime factor.
(b)
56 = 7 × 2 × 2 × 2
The numbers are prime factorised.
(c)
70 = 2 × 5 × 7
The numbers are prime factorised.
(d)
54 = 2 × 3 × 9
9 is not a prime factor.
10. Determine if 25110 is divisible
by 45. [Hint : 5 and 9 are co-prime numbers.
Test the divisibility of the number by 5 and 9].
Solution
45 = 5 × 9
As 5 and 9 are co-primes, a
number divisible by 5 and 9 will be divisible by 45.
Consider divisibility of 25110 by
5.
As the unit’s digit is 0, the
number is divisible by 5.
Consider divisibility of 25110 by
9.
The number is divisible by 9, as
the sum of the digits of the number is 9.
The number is divisible by both 5
and 9, which are co-prime numbers. Hence, the number is divisible by the
product, 45.
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.
Solution
No, not necessary.
For example, 36 is divisible by 4
and 6 but not by 24.
12. I am the smallest number, having
four different prime factors. Can you find me?
Solution
As the number is the smallest
number, it will have the smallest prime factors, 2, 3, 5 and 7.
So, 2 × 3 × 5 × 7 = 210 is the
required smallest number.
27 = 3 × 3 × 3
34 = 2 × 17
70 = 2 × 5 × 7
12 = 2 × 2 × 3
Exercise 3.6
1.
Find the HCF of the
following numbers:
The Highest Common Factor (HCF)
of two or more given numbers is the highest (or greatest) of their common
factors.
a)
18, 48
Solution
18 = 2 × 3 × 3
48 = 3 × 2 × 2 × 2 × 2
The common factors are 2 and 3.
So, HCF = 2 × 3 = 6
b)
30, 42
Solution
30 = 2 × 3 × 5
42 = 2 × 3 × 7
Common factors are 2 and 3. HCF =
2 × 3 = 6
c)
18, 60
Solution
18 = 2 × 3 × 3
60 = 2 × 2 × 3 × 5
Common factors are 2 and 3. HCF =
2 × 3 = 6
d)
27, 63
63 = 3 × 3 × 7
Common factors are 3(twice). HCF
= 3 × 3 = 9
e)
36, 84
36 = 2 × 2 × 3 × 3
84 = 2 × 2 × 3 × 7
Common factors are 2(twice) and
3. HCF is 2 × 2 × 3 = 12
f)
34, 102
34 = 2 × 17
102 = 2 × 17 × 3
Common factors are 2 and 17. HCF
= 2 × 17 = 34
g)
70, 105, 175
70 = 2 × 5 × 7
105 = 5 × 3 × 7
175 = 5 × 5 × 7
Common factors are 5 and 7. HCF =
5 × 7 = 35
h)
91, 112, 49
91 = 13 × 7
112 = 2 × 2 × 2 × 2 × 7
49 = 7 × 7
Common factor is 7. HCF = 7
i)
18, 54, 81
18 = 2 × 3 × 3
54 = 2 × 3 × 3 × 3
81 = 3 × 3 × 3 × 3
Common factors are 3 (twice). HCF
= 9
j)
12, 45, 75
45 = 5 × 3 × 3
75 = 5 × 3 × 5
Common factor is 3. HCF = 3
2.
What is the HCF of
two consecutive
i.
numbers?
Solution
The HCF of two consecutive numbers
is 1. For example, HCF of 4 and 5 is 1.
ii.
even numbers?
The HCF of two consecutive even
numbers is 2. For example, HCF of 8 and 10 is 2.
iii.
odd numbers?
The HCF of two consecutive odd
numbers is 1. For example, HCF of 13 and 15 is 1.
3.
HCF of co-prime
numbers 4 and 15 was found as follows by factorisation: 4 = 2 × 2 and 15 = 3 ×
5 since there is no common prime factor, so HCF of 4 and 15 is 0. Is the answer
correct? If not, what is the correct HCF?
Solution
The answer is incorrect.
The HCF of 4 and 15 is 1.
LCM = 36
So, when number is a factor of the other number, their LCM will be the larger number.
Exercise 3.7
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.
Solution
The required value of weight
should measure the weight of the fertiliser exact number of times. That is,
this number should be an exact divisor of the weights of the two bags. Also,
this value should be the maximum. Thus, the maximum value of weight can be
obtained by finding the HCF of the two weights, 75 kg and 69 kg.
75 = 5 × 5 × 3
69 = 23 × 3
HCF is 3.
Hence, the maximum value of
weight that can measure the weight of the fertiliser exact number of times is 3
kg.
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?
Solution
The distance covered by each one
of them is required to be the same as well as minimum. The required minimum
distance each should walk would be the lowest common multiple of the measures
of their steps.
The LCM of the three numbers is 3
× 3 × 7 × 2 × 5 × 11 = 6930 cm
So, the minimum distance each
should cover so that all can cover the distance in complete steps is 6930 cm
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.
Solution
We require the longest tape that
can measure the dimensions exactly. Thus, the maximum value can be obtained by
finding the HCF of the numbers.
825 = 5 × 5 × 3 × 11
675 = 5 × 5 × 3 × 3 × 3
450 = 5 × 5 × 3 × 3 × 2
HCF of the numbers is 5 × 5 × 3 =
75 cm
So, the longest tape that can
measure the dimensions of the room exactly is 75 cm.
4.
Determine the
smallest 3-digit number which is exactly divisible by 6, 8 and 12.
Solution
Find the LCM of the numbers 6, 8,
12.
So, LCM is 24.
But, we need to find the smallest
three-digit number.
24 × 5 = 120 is the smallest
three-digit multiple of 24. So, 120 is the smallest 3-digit number which is
exactly divisible by 6, 8 and 12.
5.
Determine the
greatest 3-digit number exactly divisible by 8, 10 and 12.
Solution
LCM = 120
Now, 120 × 8 = 960 is the largest
three-digit multiple of 120.
So, the greatest 3-digit number
exactly divisible by 8, 10 and 12 is 960.
6.
The traffic lights
at three different road crossings change after every 48 seconds, 72 seconds and
108 seconds respectively. If they change simultaneously at 7 a.m., at what time
will they change simultaneously again?
Solution
LCM = 432 seconds.
So, the traffic lights at three
different road crossings change every 432 seconds, that is, 420 minutes + 12 seconds;
7 min 12 seconds.
Hence they will change simultaneously
at 7:07:12 a.m.
7.
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 the three containers
exact number of times.
Solution
Maximum capacity is the HCF of
the numbers.
403 = 13 × 31
434 = 2 × 7 × 31
465 = 3 × 5 × 31
HCF = 31
So, the maximum capacity of the
container is 31 litres.
8.
Find the least
number which when divided by 6, 15 and 18 leave remainder 5 in each case.
Solution
Least number is 90.
Now, 90 is the least number which
leaves a remainder 0 when divided by the numbers 6, 15 and 18.
But, we need the least number
that leaves remainder 5 in each case. Therefore, the required number is 5 more
than 90. The required least number = 95.
9.
Find the smallest
4-digit number which is divisible by 18, 24 and 32.
Solution
LCM is 288
We require a 4-digit number. So,
288 × 4 = 1152 is the smallest 4-digit multiple of 288.
So, 1152 is the required number.
10. 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 LCMs. Is LCM the product of two numbers in each case?
Solution
LCM = 60
LCM = 30
LCM = 60
We see that the LCM of the
numbers is the products of the given numbers. This is because the given numbers
are co-primes.
Thus, when the numbers are
co-primes, the LCM of the numbers is their product.
11. 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?
Solution
LCM = 20
LCM = 18
LCM = 48
LCM = 45
We
see that the LCM of the numbers is the larger number.
So, when number is a factor of the other number, their LCM will be the larger number.
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