BATRA COACHING CENTRE
1326/14, Arjun Nagar, Rohtak, : 9416124333
REAL NUMBERS
DIVISIBILITY
A non-zero integer ‘a’ is said to divide an integer ‘b’ if there exists an integer c such that
b = ac.
The integer ‘b’ is called the dividend, integer ‘a’ is known as the divisor and integer ‘c’ is
known as the quotient.
For example, 4 divides 48 because there is an integer 12 such that 48 = 4  12. However,
4 does not divide 35 because there do not exist an integer c such that 35 = 4  c. In other words
35 = 4  c is not true for any integer c.
If a non-zero integer ‘a’ divides an integer b, then we write ab. This is read as “a divides
b”. When ab, we say that ‘b is divisible by a’ or ‘a is a factor of b’ or ‘b is a multiple of a’ or ‘a is
a divisor of b’.
ILLUSTRATION State whether the following are true or not:
(i)
393
(ii)
04
(iii)
50
(iv)
-28
(v)
8-8
(vi)
13 -25
(vii) 1-1
Solution: We observe that:
(i)
393 is true, because 93 = 3  
(ii)
04 is not true by definition.
(iii)
5  0 is true, because 0 = 5  0
(iv)
-28 is true, because 8 = (-2)  (-4)
(v)
8-8 is true, because – 8 = 8  -1
(vi)
13-25 is not true, because -25 = 13 c is not valid for any integer c.
(vii) 1-1 is true, because -1 = 1 
IMPORTANT PROPERTIES
(i)
1 divides every non-zero integer.
i.e., 1 a for every non-zero integer a.
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
0 is divisible by every non-zero integer a.
i.e.,a 0 for every non-zero integer a.
0 does not divide any integer.
If a is a non-zero integer and b is any integer, then
ab  a-b, -ab and -a-b
If a and b are non-zero integers, then
ab and ba  a = b
If a is a non-zero integer and b, c are any two integers, then
a b  c

ab and ac  a bc

for any integer x.
a bx
If a and c are non-zero integers and b, d are any two integers, then
(i)
ab and cd acbd
1
(ii)
acbc  ab
IMPORTANT NOTE
Whenever we will speak of divisors in this chapter we will mean positive divisors of
positive integers.
EUCLID’S DIVISION LEMMA
This lemma is nothing but a restatement of the long division process we have been
doing for the last many years.
Consider the division of one positive integer by another, say 71 by 9. The division can be
carried out as follows:
9 71 7
63
8
So
71 = 9  + 8, 0  8 < 9
SOME MORE EXAMPLES
Pair of integers
Representation
25, 7
25 = 7  3 + 4, 0  4 < 7
20,3
20 = 3  6 + 2, 0  2 < 3
7,15
7 = 150+7, 0  7 < 15
35,5
35 = 57 + 0, 0  0 < 5
EUCLID’S DIVISION LEMMA
Let a and b be any two positive integers. Then, there exists unique integers q and r such
that
a = bq + r, 0  r < b
If ba, then r = 0, Otherwise, r satisfies the inequality 0 < r < b.
ILLUSTRATIVE EXAMPLE
Q.1
Show that every positive even integer is of the form 2q, and that every positive odd
integer is of the form 2q + 1, where q is some integer.
Sol.
Let a be any positive integer and b = 2. Then, by Euclid’s division Lemma there exist
integers q and r such that
a = 2q + r, where 0  r < 2
Now,
0  r < 2  0  r  1  r = 0 or, r = 1

a = 2q or, a = 2q + 1
If a = 2q, then a is an even integer.
We know that an integer can be either even or odd. Therefore, any odd integer is of the form
2q + 1.
Q.2
Show that any positive integer is of the form 3q or, 3q + 1 or , 3q + 2 for some integer q.
Sol.
Let a be any positive integer and b = 3. Applying division Lemma with a and b= 3, we
have
a = 3q + r, where 0  r < 3 and q is some integer
2

r 

a = 3q or, a = 3q + 1 or, a = 3q + 2 for some integer q.
Q.3
Show that n2 – 1 is divisible by 8, if n is an odd positive integer.
Sol. We know that any odd positive integer is of the form 4q + 1 or, 4q + 3 for some integer q.
So, we have the following cases:
Case I When n = 4q + 1
In this case, we have
n2 -1 = (4q + 1)2 – 1 = 16q2 + 8q + 1 – 1 = 16q2 + 8q = 8q (2q + 1)

n2 – 1 is divisible by 8.
Case II When n = 4q + 3
In this case, we have
n2 – 1 = (4q + 3)2 – 1 = 16q2 + 24q + 9 – 1 = 16q2 + 24q + 8

n2 – 1 = 8(2q2 + 3q + 1)

n2 – 1 is divisible by 8
2
Hence, n – 1 is divisible by 8.
Q.4
Prove that if x and y are odd positive integers, then x2 + y2 is even but not divisible by 4.
Sol.
We know that any odd positive integer is of the form 2q + 1 for some integer q.
So, let x = 2m + 1 and y = 2n + 1 for some integers m and n.

x2 + y2 = (2m + 1)2 + (2n + 1)2

x2 + y2 = 4m2+ 4m + 1 + 4n2 + 4n + 1

x2 + y2 = 4(m2 + n2) + 4(m + n) + 2

x2 + y2 = 4 {(m2 + n2) + (m + n)} + 2

x2 + y2 = 4q + 2, where q = (m2 + n2) + (m + n)

x2 + y2 is even and leaves remainder 2 when divided by 4

x2 + y2 is even but not divisible by 4.
Q.5
Use Euclid’s division Lemma to show that the cube of any positive integer is either of the
form 9m, 9m + 1 or, 9m + 8 for some integer m.
Sol.
Let x be any positive integer. Then, it is of the form 3q or 3q + 1 or 3q + 2. So, we have
the following cases:
Case I When x = 3q
In this case, we have
x3 = (3q)3 = 27q3 = 9(3q3) = 9m, where m = 3q3
Case II When x = 3q + 1
In this case, we have
x3 = (3q + 1)3

x3 = 27q3 + 27q2 + 9q + 1

x3 = 9q (3q2 + 3q + 1) + 1

x3 = 9m + 1, where m = q(3q2 + 3q + 1)
Case III When x = 3q + 2
In this case, we have
x3 = (3q + 2)3

x3 = 27q3 + 54q2 + 36q + 8

x3 = 9q (3q2 + 6q + 4) + 8
3

x3 = 9m + 8, where m = q (3q2 + 6q + 4)
Hence, x3 is either of the form 9m or, 9m + 1 or 9m + 8.
Q.6
Show that one and only one out of n, n + 2 and n + 4 is divisible by 3, where n is any
positive integer.
Sol.
We know that any positive integer is of the form 3q or 3q + 1 or 3q + 2 for some integer
q and one and only one of these possibilities can occur.
So, we have following cases :
Case I When n = 3q
In this case, we have
n = 3q, which is divisible by 3
Now, n = 3q

n + 2 = 3q + 2,

n + 2 leaves remainder 2 when divided by 3

n + 2 is not divisible by 3
Again, n = 3q

n + 4 = 3q + 4 = 3 (q + 1) + 1

n + 4 leaves remainder 1 when divided by 3

n + 4 is not divisible by 3
Thus, n is divisible by 3 but n + 2 and n + 4 are not divisible by 3.
Case II When n = 3q + 1
In this case, we have
n = 3q + 1

n leaves remainder 1 when divided by 3

n is not divisible by 3
Now, n = 3q + 1

n + 2 = (3q + 1) + 2 = 3 (q + 1)

n + 2 is divisible by 3
Again, n = 3q + 1

n + 4 = 3q + 1 + 4 = 3q + 5 = 3 (q + 1) + 2

n + 4 leaves remainder 2 when divided by 3

n + 4 is not divisible by 3
Thus, n + 2 is divisible by 3 but n and n + 4 are not divisible by 3
Case III When n = 3q + 2
In this case, we have
n = 3q + 2

n leaves remainder 2 when divided by 3

n is not divisible by 3
Now, n = 3q + 2

n + 2 = 3q + 2 + 2 = 3(q + 1) + 1

n + 2 leaves remainder 1 when divided by 3

n + 2 is not divisible by 3
Again, n = 3q + 2
n + 4 = 3q + 2 + 4 = 3 (q + 2)
4

n + 4 is divisible by 3
Thus, n + 4 is divisible by 3 but n and n + 2 are not divisible by 3.
Q.7
Prove that one of every three consecutive positive integers is divisible by 3.
Sol.
Let n, n + 1, n + 2 be three consecutive positive integers.
We know that n is of the form 3q, 3q + 1 or, 3q + 2.
So, we have the following cases:
Case I When n = 3q
In this case, n is divisible by 3 but n + 1 and n + 2 are not divisible by 3.
Case II When n = 3q + 1
In this case, n + 2 = 3q + 1 + 2 = 3(q + 1) is divisible by 3 but n and n + 1 are not divisible by 3.
Case III When n = 3q + 2
In this case, n + 1 = 3q + 1 + 2 = 3(q + 1) is divisible by 3 but n and n + 2 are not divisible by 3.
Hence, one of n, n + 1 and n + 2 is divisible by 3.
Q.8
Show that the square of any positive integer is of the form 3m or 3m + 1, for some
integer m.
Sol.
Let x be a positive integer

x = 3q, 3q + 1 or 3q + 2
2
If x = 3q, then
x = 9q2 = 3(3q2) = 3m
If x = 3q + 1, then
x2 = (3q + 1)2 = 9q2 + 6q + 1 = 3(3q2 + 2q) + 1
= 3m + 1 for some integer m
2
If x = 3q + 2, then
x = (3q + 2)2 = 9q2 + 12q + 4 = 3(3q2 + 4q + 1) + 1
= 3m + 1 for some integer m.
Q.9
Show that the square of any positive odd integer is of the form 8m + 1, for some integer
m.
Sol.
Every positive odd integer n is of the form 4q + 1 or 4q + 3, for some integer q.
Case 1 When n = 4q + 1,
n2 = (4q + 1)2 = 16q2 + 8q + 1 = 8(2q2 + q) + 1
= 8m + 1, where m = 2q2 + q
Case 2 When n = 4q + 3,
n2 = (4q + 3)2 = 16q2 + 24q + 9 = 8(2q2 + 3q + 1) + 1
= 8m + 1, where m = 2q2 + 3q + 1
Hence, the square of any positive odd integer is of the form 8m + 1, for some integer m.
Q.10 Prove that if a positive integer is of the form 6q + 5, then it is of the form 3m + 2 for
some integer m but not conversely.
Sol.
The given integer = 6q + 5 = 3  2q + 3 + 2 = 3  (2q + 1) + 2
= 3m + 2, where m = 2q + 1.
Conversely, consider the integer 8.
Now,
8 = 6 + 2 = 32, i.e., it is of the form 3m + 2.
Suppose
8 = 6q + 5

3 = 6q  2q = 1
Which has no solution in integers.
 8 is not of the form, 6q + 5.
EXERCISE 1
5
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
Prove that the square of any positive integer is of the form 4q or 4q + 1 for some integer
q.
Prove that the square of any positive integer is of the form 5q, 5q + 1, 5q + 4 from some
integer q.
Prove that if a positive integer is of the form 6q + 5, then it is of the form 3q + 2 for
some integer q, but not conversely.
Prove that the square of any positive integer of the form 5q + 1 is of the same form.
Prove that the product of three consecutive positive integer is divisible by 6.
For any positive integer n, prove that n3 – n divisible by 6.
Show that any positive integer is of the form 3q or 3q + 1 or 3q + 2 for some integer q.
Show that an odd integer is of the forms 6q + 1, 6q + 3 or 6q + 5, where q is some
integer.
Using Euclid’s Division Lemma, show that the cube of any positive integer is of the form
9p, 9p + 1 or 9p + 8, where p is some integer.
Prove that n2 – n is even for every positive integer n.
If m and n (m > n) be two odd positive integers, then prove that one of the numbers
mn
mn
and
is odd and the other is even.
2
2
EXERCISE 1 - HINTS AND SOLUTION
1.
Since any positive integer n is of the form 2m or, 2m + 1
If n = 2m, then
n2 = 4m2 = 4q, where q = m2
If n = 2m + 1, then
n2 = (2m + 1)2 = 4m2 + 4m + 1 = 4m(m + 1) + 1 = 4q + 1, where q = m (m + 1)
2.
Since any positive integer n is of the form 5m or 5m + 1, or 5m + 2 or 5m + 3 or 5m + 4.
If n = 5m, then
n2 = 25m2 = 5(5m) = 5q, where q = 5m
If n = 5m + 1, then
n2 = (5m + 1)2
= 25m2 + 10m + 1
= 5m(5m + 2) + 1 = 5q + 1, where q = m (5m + 2)
If n = 5m + 2, then
n2 = (5m + 2)2 = 25m2 + 20m + 4
= 5m (5m + 4) + 4 = 5q + 4, where q = m (5m + 4)
If n = 5m + 3, then
n2 = (5m + 3)2 = 25m2 + 30m + 9 = 25m2 + 30m + 5 + 4
= 5(m2 + 6m + 1) + 4 = 5q + 4, where q = 5m2 + 6m + 1
If n = 5m + 4, then
n2 = (5m + 4)2 = 25m2 + 40m + 16 = 25m2 + 40m + 15 + 1
= 5(5m2 + 8m + 3) + 1 = 5q + 1, where q = 5m2 + 8m + 3
Hence, n2 is of the form 5q or, 5q + 1 or, 5q + 4.
3.
Let n = 6q + 5, where q is a positive integer. We know that any positive integer is of the
form 3k or, 3k + 1 or, 3k + 2
6
4.
5.
6.
9.
10.
11.

q = 3k or, 3k + 1 or, 3k + 2
If q = 3k, then
n = 6q + 5 = 18k + 5 = 3(6k + 1) + 2 = 3m + 2, where m = 6k + 1
If q = 3k + 1, then
n = 6q + 5 = 6(3k + 1) + 5 = 18k + 6 + 5 = 18k + 11 = 18k + 9 + 2
= 3 (6k + 3) + 2 = 3m + 2, where m = 6k + 3
If q = 3k + 2, then
n = 6q + 5 = 6 (3k + 2) + 5 = 18k + 12 + 5 = 18k + 17 = 18k + 15 + 2
= 3 (6k + 5) + 2 = 3m + 2, where m = 6k + 5.
Let n = 5q + 1. Then,
n2 = 25q2 + 10q + 1 = 5(5q2 + 2q) + 1 = 5m + 1, where m = 5q2 + 2q

n2 is of the form 5m + 1.
Let n be any positive integer. Since any positive integer is of the form 6q, 6q + 1 , 6q + 2,
6q + 3, 6q + 4, or 6q + 5
If n = 6q, then
n(n + 1) (n + 2) = 6q (6q + 1) (6q + 2), which is divisible by 6
If n = 6q + 1, then
n(n + 1) (n + 2) = (6q + 1) (6q + 2) (6q + 3) = 6(6q + 1) (3q + 1) (2q + 1),
Which is divisible by 6.
If n 6q + 2, then
n(n + 1) (n + 2) = (6q + 2) (6q + 3) (6q + 4) = 12 (3q + 1) (2q + 1) (2q + 3).
Which is divisible by 6.
Similarly, n(n + 1) (n + 2) is divisible by 6 if n = 6q + 3 or, 6q + 4 or, 6q + 5.
We have,
n3 – n = (n – 1) (n) (n + 1), which is product of three consecutive positive integers.
So, proceed as in Q. No. 10.
Hint: Any positive integer n can be of the form 3q, 3q + 1 or 3q + 2.
Hint: Any positive integer n can be of form 2k or 2k + 1.
mn
mn
 k  l  1,
 k–l
Hint: Take m = 2k + 1 and n = 2l + 1. As m > n  k > 1 
2
2
If k – l is odd, then k + l + 1 = k – l + (2l + 1) is even and vice versa.
7
EXAMPLES
Q.1
Use Euclid’s division algorithm to find the HCF of 210 and 55.
Sol.
210 = 55  3 + 45
….. (i)
55 = 45 


…..(ii)
45 = 10  4 + 5
….. (iii)
10 = 5  2 + 0
….. (iv)
The remainder at this stage is zero. So, the divisor at this stage or the remainder at the previous
stage is 5 i.e. 5 is the HCF of 210 and 55.
Q.2
Use Euclid’s division algorithm to find the HCF of 4052 and 12576.
Sol.
12576 = 4052  3 + 420
…..(i)
4052 = 420  9 + 272
…..(ii)
420 = 272  1 + 148
….(iii)
272 = 148  1 + 124
….. (iv)
148 = 124  1 + 24
…..(v)
124 = 24 5 + 4
…..(vi)
24 = 4  6 + 0
…..(vii)
We observe that the remainder at this stage is zero. Therefore, the divisor at this stage is 4(or
the remainder at the earlier stage) is the HCF of 4052 and 12576.
Q.3
Find the HCF of 81 and 237 and express it as a linear combination of 81 and 237.
Sol.
237 = 81  2 + 75
….(i)
81 = 75  1 + 6
….(ii)
75 = 6  12 + 3
…..(iii)
We consider the new divisor 6 and the new remainder 3 and apply division lemma to get
6=32+0
The remainder at this stage is zero. So, the divisor at this stage or the remainder at the earlier
stage i.e. 3 is the HCF of 81 and 237.
To represent the HCF as a linear combination of the given two numbers, we start from the last
but one step and successively eliminate the previous remainders as follows:
From (iii), we have
3 = 75 - 612



3 = 75 – 12  81 + 12  75

3 = 13 75 – 12  81







 237x + 81y, where x = 13 and y = -38
NOTE:- It follows from the above example that the HCF (say d) of two positive integers a and b
can be expressed as a linear combination of a and b i.e., d = xa + yb for some integers x and y.
Also, this representation is not unique.
Q.4
Find the HCF of 65 and 117 and express it in the form 65 m + 117 n.
Sol.
Given integers are 65 and 117 such that 117 > 65.
Applying division lemma to 65 and 117, we get
8
117 = 65  1 + 52
……(i)
65 = 52  1 + 13
……(ii)
52 = 13  3 + 0
…..(iii)
13 is the HCF of 65 and 117.
From (ii), we have
13 = 65 – 52  1

13 = 65 –(117 – 65  1)

13 = 65 – 117 + 65  1

13 = 65  2 + 117  (-1)

13 = 65 m + 117n, where m = 2 and n = -1.
Q.5
If the HCF of 210 and 55 is expressible in the form 210  5 + 55y, find y.
Sol.
210 = 55 


(i)
55 = 45  1 + 10
……(ii)
45 = 4  10 + 5
…..(iii)
10 = 5  2 + 0
…..(iv)
We observe that the remainder at this stage is zero. So, the last divisor i.e. 5 is the HCF
of 210 and 55.

5 = 210  5 + 55y

55y = 5 – 210  5 =5-1050

55y = -1045
 1045
y
 19

55
Q.6
A sweet seller has 420 Kaju burfis and 130 Badam burfis she wants to stack them in such
a way that each stack has the same number, and they take up the least area of the tray.
What is the number of burfis that can be placed in each stack for this purpose?
Sol.
The area of the tray that is used up in stacking the burfis will be least if the sweet seller
stacks maximum number of burfis in each stack. Since each stack must have the same
number of burfis. Therefore, the number of stack will be least if the number of burfis in
each stack is equal to the HCF of 420 and 130.
In order to find the HCF of 420 and 130, let us apply Euclid’s division lemma to 420 and
130 to get
420 = 130  3 + 30
……(i)
130 = 30  4 + 10
……(ii)
30 = 3  10 + 0
……(iii)
Since the remainder at this stage is zero. Therefore, last divisor 10 is the HCF of 420 and
130.
Hence, the sweet seller can make stacks of 10 burfis of each kind to cover the least area
of the tray.
Q.7
An army contingent of 616 members is to march behind an army band of 32 members in
a parade. The two groups have to march in the same number of columns. What is the
maximum number of columns in which they can march?
Sol.
Let n be the maximum number of columns in which the two groups can march.
Then, n is the HCF of 616 and 32, i.e.
9
HCF (616, 32) = n
616 = 32 
32 = 8 4 + 0
Hence,
HCF (616, 32 ) = HCF (32, 8) = 8
Hence, the maximum nu8mber of columns in which they can march is 8.
Q.8
Find the largest number that will divide 398, 436 and 542 leaving remainders 7, 11 and
15 respectively.
Sol.
Clearly, the required number is the HCF of the numbers
398 – 7 = 391, 436 – 11 = 425, and, 542 – 15 = 527.
First we find the HCF of 391 and 425 by Euclid’s algorithm
425 = 391  1 + 34
391 = 34  11 + 17
34 = 17  2 + 0
 HCF of 391 and 425 = 17
Now we have to find HCF of 527 and 17
527 = 17  31 + 0
 HCF = 17
So required numbers = 17
Q.9
Two tankers contain 850 litres and 680 litres of petrol respectively. Find the maximum
capacity of a container which can measure the petrol of either tanker in exact number
of times.
Sol.
Clearly, the maximum capacity of the container is the HCF of 850 and 680 in litres.
850 = 680  1 + 170
680 = 170  4 + 0
 HCF = 170
 Capacity of container = 170 Litre.
ASSIGNMENT
1.
Define HCF of two positive integers and find the HCF of the following pairs of numbers:
(i)
18 and 24
(ii)
75 and 243
(iii)
155 and 1385
2.
Use Euclid’s division algorithm to find the HCF of
(i)
135 and 225
(ii)
196 and 38220
(iii)
867 and 255.
3.
Find the HCF of the following pairs of integers and express it as a linear combination of
them.
(i)
592 and 252
(ii)
1288 and 575.
4.
Find the HCF and LCM of the given integers, using the prime factorization method :
(i)
8, 12 and 20
(ii)
50, 160 and 400
(iii)
40, 110 and 360
(iv)
80, 90 and 250
5.
Find the LCM of 1160 and 1240, if HCF of 1160 and 1240 is 40.
6.
If the HCF of 408 and 1032 is expressible in the form 1032 m – 408  5, find m.
7.
If the HCF of 657 and 963 is expressible in the form 657x + 963 (-15), find x.
8.
Find the greatest number which divides 285 and 1249 leaving remainders 9 and 7
respectively.
10
9.
Find the largest number which exactly divides 280 and 1245 leaving remainders of 4 and
3 respectively.
10.
Find the greatest number which divides 2011 and 2623 leaving remainders 9 and 5
respectively.
11.
The length, breadth and height of a room are 8m 25 cm, 6m 75 cm and 4m 50 cm,
respectively. Determine the longest rod which can measure the three dimensions of the
room exactly.
12.
15 pastries and 12 biscuit packets have been donated for a school fete. These are to be
packed in several smaller identical boxes with the same number of pastries and biscuit
packets in each. How many biscuit packets and how many pastries will each box
contain?
13.
144 cartons of Coke Cans and 90 cartoons of Pepsi Cans are to be stacked in a Canteen.
If each stack is of the same height and is to contain cartoons of the same drink, what
would be the greatest number of cartoons each stack would have?
14.
A sweetseller has 200 Kaju burfis and 75 badam burfis. He wants to stack them in such a
way that each stack has the same number, and they take up the least area of the tray.
What is the maximum number of burfis that can be placed in each stack for this
purpose?
Fundamental Theorem of Arithmetic (unique Factorisation Theorem)
Theorem
Every natural number greater than one either is itself a prime number or can be
written as a product of prime numbers in a unique way.
Or
Every composite number can be expressed (factorised) as a product of primes,
and this factorization is unique, apart from the order in which the prime factors
occur.
Applications of Fundamental Theorem of Arithmetic
The Fundamental Theorem of Arithmetic has several applications.
Q.1
Express each of the following positive integer as the product of its prime factors:
(i)
140
(ii)
3825
(iii)
5005
Sol.
(i)
Q.2
(i)
Sol.
140 = 22  5  7.
(ii)
3825 = 32  52  17. (iii)
5005 = 5  7  11 
Determine the prime factorization of each of the following positive integers:
10350
(ii)
27720
11
 Their prime factorizations are
(i)
10350 = 2 3  3  5  5  23
= 2  32  52  23.
(ii)
27720 = 2  2  2  3  3  5  7  11
= 23   11.
Q.3
Find the LCM and HCF of 14 and 40 by prime factorization method.
Sol.
We know that
14 = 2  7
30 = 2 
The product of the smallest power of each common prime factor gives the HCF, and the
product of the greatest power of each prime factors gives the LCM.
Thus, HCF (14, 40) = 21 = 2
and LCM (14, 40) = 23  5  7 = 280.
NOTE: If a and b are any two positive integers, then
HCF (a, b)  LCM (a, b ) = a  b.
With the help of this property, we can easily find the LCM of two large numbers conveniently.
This property, however, does not hold true in case of three or more integers.
Q.4
Find the LCM of 30240 and 29700.
Sol.
The prime factorizations of 30240 and 29700 are
30240 = 25  33  5  7
29700 = 22  33  52  7
HCF (30240, 29700) = 22  32  5 = 540.
Using the Property stated above, we have
30240  29700
 1663200.
LCM (30240, 29700) =
540
Q.5
Find the HCF and LCM of 40, 125 and 280, using the prime factorization method.
Sol.
Here,
40 = 23  5
125 = 53
280 = 23  5  7
Thus, HCF (40, 125, 280) = 5
and LCM (40, 125, 280) = 23  53  7 = 7000.
Note that
12
Q.6
Sol.
Q.7
Sol.
Q.8
Sol.





Q.9
Sol.
HCF  LCM  Product of numbers
(5  7000)  (40  125  280).
Find the largest positive integer that will divide 434 and 539 leaving remainders 9 and
12 respectively.
It is given that the number 434 leaves remainder 9 and the number 539 leaves the
remainder 12.
 434 – 9 = 425 and 539 – 12 = 527 are divisible by the required largest positive integer.
Thus we need to find HCF of 425 and 527.
Now
425 = 5 
and
527 = 17  31.

HCF (425, 527) = 17
 The required integer = 17.
Explain why 7 11  13 + 13 and 7  6  5  4  3  2  1 + 5 are composite numbers.
Let
a = 7  11  13 
= [7  11 + 1]  13
= 78  13
Hence, 13 is a factor of a
 a is a composite number.
Similarly, let
b =7654321+5
=7643215+5
= [7  6  4  3  2  1 + 1]  5
= (1008 + 1)  5
= 1009  5
Hence, 5 is a factor of b.
Hence, b is a composite number.
Check whether 6n can end with the digit 0 for any n  N.
Let, if possible, the composite number 6n have a value which ends with the digit 0.
10 is a factor of 6n
6n = 10  q, where q is some natural number
(2  3)n = 2  5  q

2n  3n = 2  5  q
The assumption 5 is a prime factor of 2n  3n is not possible because 2n  3n can have
only 2 and 3 as prime factors. Hence, our assumption is wrong.
Thus, 6n cannot end with the digit 0.
There is a circular path around a sports field. Sonia takes 18 minutes to drive one round
of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the
same point and at the same time, and go in the same direction, after how many minutes
will they meet again at the starting point?
LCM of 18 minutes and 12 minutes is 36 minutes.
In 36 minutes, Sonia will arrive at the starting point after making 2 rounds, while in 36
minutes, Ravi will arrive at the same starting point after making 3 rounds.
Hence, they meet each other at the starting point after 36 minutes.
13
Q.10
Sol.
In a seminar, the number of participants in Urdu, Hindi and English are 45, 75 and 135
respectively. Find the number of rooms required to house them if in each room, the
same number of participants are to be accommodated and all of them must belong to
the same language.
Since in each room, the same number of participants, belonging to the same language
are to be accommodated, their number in each room must be HCF of 45, 75 and 135.
Now
45 = 32  5, 75 = 3  52 and 135 = 32  5
 HCF (45, 75, 135) = 3  5 = 15.
 Each room accommodates 15 participants.
 Total number or rooms required
Total number of participan ts
=
15
45  75  135 255

 17.
=
15
15
Exercise
Q.1
Find the prime factorization of the following:
(i)
84
(ii)
120
(iii)
256 (iv)
760 (v)
8008
Q.2
Find the HCF and LCM of the following integers, using the Fundamental Theorem of
Arithmetic:
(i)
60 and 75
(ii)
105 and 240
(iii)
1176 and 6384
(iv)
2448 and 4284
Q.3
The HCF of two numbers is 54 and their LCM is 3024. If one number is 336, find the
other number.
Q.4
Can two numbers have 18 as their HCF and 694 as their LCM ? Give reason.
Q.5
Find the least number that is divisible by all the number between 2 and 10 (both
inclusive).
Q.6
Check whether 4n, where n is a natural number, ends with the digit zero.
Q.7
A circular field has a circumference of 360 km. Three cyclists start together and can cycle
48, 60 and 72 km a day, round the field. When will they meet again?
Q.8
In a government girls school, there are two sections in Class X, Section A and Section B
which respectively have 40 and 48 girls. Find the minimum number of books required
for their class-library so that they can be distributed equally among the students of
Sections A and B.
Q.9
Using prime factorization method, find the HCF and the LCM of 72, 126 and 168. Also,
show that HCF  LCM = Product of the three numbers.
Q.10 Find the [HCF  LCM] for the numbers 100 and 190.
Example
Q.1
Prove that 2 is an irrational number.
Sol.
We shall prove this by contradiction method.
Let us assume, to the contrary, that 2 is a rational number.
14
p
, where p and q are integers having no common factor, and q > 0.
q
On squaring both the sides, we have
p2
2  2  p 2  2q 2
q

2 divides p2
it follows that 2 divides p.
So, we have p = 2m, for some integer m,
Substituting for p in Eq. (1), we get
(2m)2 = 2q2

q2 = 2m2
Arguing as above, we get 2 divides q.
From Eq. (2) and Eq. (3), we get that p and q have 2 as a common factor. But this
contradicts the fact that p and q have no common factor. This contradiction implies that
2 is not a rational number, i.e., 2 is an irrational number.
Prove that 3 is an irrational number.

Q.2
Sol.
Let us assume 3 is a rational number.
p
Then 3  , where p and q are integers having no common factor, and q > 0.
q
p2
q2

3q2 = p2

Since 3 divides 3q2, 3 also divides p2

3 divides p

p = 3n, where n is an integer.
Also, 3q2 = p2

3q2 = (3n)2 = 9n2

q2 = 3n2
Arguing as above, we get 3 divides q
Thus, both p and q have a common factor 3 which is a contradiction.
Hence, 3 is an irrational number.
Prove that there is no rational number whose square is 6.
a
Let be a rational number whose square is 6.
b

Q.3
Sol.
2
3
2
a
6 =   , where a and b are prime to each other and b > 0
b
2

6b = a2

3(2b2) = a2
2
Thus, a is divisible by 3, and so a is divisible by 3.
Let a = 3p, where p is some integer.
Then
15
a2 = 6b2 gives
(3p)2 = 6b2

9p2 = 6b2

3p2 = 2b2
This shows that 2b2 is divisible by 3, so b2 is divisible by 3, and thus b is divisible by 3.
Thus, both a and b are divisible by 3 which is a contradiction.
Hence, 6 is not a rational number.
Show that 3 2 is an irrational number.
Let us assume, that 3 2 is a rational number.
a

3 2  , where a and b are integers with no common factor and b > 0.
b
a
2

3b
a
Since a and b are integers,
is rational, and so 2 is rational.
3b
But this contradicts the fact that 2 is an irrational number.
Hence, 3 2 is an irrational number.
Show that 3  2 5 is an irrational number.
Let us assume, that 3  2 5 is a rational number. Then there exist co-prime positive
integers a and b such that
a
3 2 5 
b
a
2 5  3

b
a  3b
2 5

b
a  3b
5

2b
a  3b
Since a and b are integers,
is a rational number; and hence 5 is rational. But
2b
this contradicts the fact that 5 is irrational.
Hence, 3  2 5 is an irrational number.
Prove that 3  7 is an irrational number.
Let us assume, that 3  7 is a rational number.
p
Then 3  7  where p, q are integer having no common factor and q > 0.
q
p
 7  3

q
Then,
Q.4
Sol.
Q.5
Sol.
Q.6
Sol.
16
2




p

7    3 
q

2
p
2p
7  2 3
3
q
q
2p
p2
p 2  4q 2
3  2 4 
q
q
q2
3
p 2  4q 2
2 pq
[ p, q are int egers, so
p 2  4q 2
is rational .]
2 pq
 3 is a rational number.
This contradicts the fact that 3 is irrational.
Hence, our assumption is wrong.
Thus, 3  7 is an irrational number.
Exercise
Q.1
Prove that each of 5 and 7 is an irrational number.
Q.2
Prove that each of the following is an irrational number.
1
3 7
(i)
(ii)
(iii)
(iv)
2 11
3  2 (v)
3 5 2
3
Q.3
Prove that each of the following is an irrational number.
4
52 3
5 7
(i)
(ii)
(iii)
5
Q.4
Prove that 7  3 2 is not a rational number.
Q.5
Show that there is no positive integer n for which n  1  n  1 is rational.
p
q
1
[Hint: Assume, n  1  n  1 is rational, say n  1  n  1  , then 
q
p
n 1  n 1
2q
n 1  n 1 
. Add and subtract to get the required contradiction.]

q
Revisiting Decimal Representation of Rational Number
Theorem 1.4 Let x be a rational number whose decimal expansion terminates. Then x can be
p
expressed in the form , where p and q are co-prime, and the prime factorization of q is of the
q
m
n
form 2  5 , where m and n are non-negative integers.
p
Theorem 1.5 (Converse of Theorem 1.4) Let x =
be a rational number, such that the prime
q
factorization of q is of the form 2m  5n, where m and n are non-negative integers. Then x has a
decimal expansion which terminates.
Examples.
2 2 2 2 4
17
17
17  5
85
 1

 0.4
 2 1  2 2  2  0.85
(i)
(ii)
5 5
2  5 10
20 2  5
2 5
10
17
(iii)
(iv)
31
31
31 5 155
 3 2  3 3  3  0.155
200 2  5
2 5
10
4
91 91 91 2
9116 1456
 4  4 4 

 0.1456
625 5
2 5
10 4
10 4
Note: Here, we have converted a rational number of the form
p
, where q is of the form
q
a
, where b is a power of 10. Therefore,
b
the decimal expansion of such a rational number terminates.
p
Theorem 1.6 Let x =
be a rational number such that p and q are co-prime and the prime
q
factorization of q is not of the form 2m  5n, where m and n are non-negative integers. Then x
has a decimal expansion which is non-terminating repeating.
Examples.
64
64
64

. So,
(i)
has a non-terminating repeating decimal expansion.
455 5  7 13
455
29 29
29
 3 . So,
(ii)
has a non-terminating repeating decimal expansion.
343 7
343
77 11
11
77


. So,
(iii)
has a non-terminating repeating decimal expansion.
210 30 2  3  5
210
Note: That the denominators also have factors other than 2, 5.
Exercise
1.
Without actually performing division, state whether the following rational number will
have a terminating decimal expansion or a non-terminating repeating decimal
expansion.
109
51
11
137
(i)
(ii)
(iii)
(iv)
125
260
3125
1500
15
217
23
313
(v)
(vi)
(vii)
(vii)
10625
4000
6250
3500
3103
1115
721
215
(ix)
(x)
(xi)
(xii)
2916
10125
2520
4500
2.
What can you say about the prime factorization of the denominators of the following
rational numbers ?
(i)
37.3507
(ii)
(iii)
272.25123
(iv)
410.37125
572.1234569
[Hints: (i) is a terminating decimal expansion. So, its denominator is of the form 2m  5n, where
m and n are non-negative integers.
(ii) Since 572.1234569 has non-terminating repeating decimal expansion, its
denominator has factors other than 2 and 5.]
Exercise
Answer each of the following questions either in one word or one sentence or as per the
requirement of the question.
1.
Write 546 as product of its prime factors.
2m  5n, to an equivalent rational number of the form
18
2.
3.
5.(i)
(ii)
Write a rational number between 2 and 3.
Write an irrational number between 1 and 2.
3
Write an irrational number between
and 5.
2
The LCM of two number is 119 and their product is 4284. Find their HCF.
Find the [HCF  LCM] for the numbers 105 and 120.
6.
Write the condition to be satisfied by q so that rational number
4.
7.
8.
9.
10.
p
has a nonq
terminating decimal expansion.
Write the exponent of 5 in the prime factorization of 3500.
Write the sume of the exponents of prime factors in the prime factorization of 84.
p
If
is a rational number (q  0), what is the condition on q so that the decimal
q
p
representation of
is terminating?
q
43
The decimal expansion of the rational number 4 3 will terminate after how many
2 .5
places of decimal?
19