08 - Question bank - Real Numbers - Class 10 - Maths Class 10 Notes | EduRev
Crash Course for Class 10 Maths by Let`s tute
Class 10 : 08 - Question bank - Real Numbers - Class 10 - Maths Class 10 Notes | EduRev
Page 1
REAL
EUCLID’S DIVISION ALGORITHM
1) Use Euclid’s division algorithm to find the HCF of 1362 and 6954
Solution:
STEP 1: Since 6954>1362 we apply division lemma to 6954 and 1362 to get,
6 9 5 4 1 3 6 2 x 5 1 4 4 = +
STEP 2: Since the remainder 144 0
1 3 6 2 1 4 4 x 9 + 6 6
STEP 3: Since the remainder
1 4 4 6 6 x 2 + 1 2 =
STEP 4: Since the remainder
6 6 1 2 x 5 + 6 =
STEP 5: Since the remainder
1 2 6 x 2 + 0 =
The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 6, the
HCF of 1362 and 6954 is 6.
2) Show that any positive odd integer is of the form
Solution:
Let’s start with taking a, where a is a positive odd integer. We apply the division algorithm with
and b =8
Since 0 8 r = < the possible remainders are
That is, a can be 8q or 8q+1 or 8q+2 or 8
However, since a is odd a cannot be 8
Therefore any odd integer can only be of the form
3) A fruit seller has 56 red apples and
has the same number, and they take up the least area of the tray.
that can be placed in each stack for this purpose?
Solution:
To find the solution we have to first find the HCF of 56 and 32
maximum number of apples in each stack
take up least amount of space.
We use Euclid’s division algorithm to find the HCF of 56 and 32
56 = 32 x 1 + 24
Since, 24 0 ? ? 32 = 24 x 1 + 8
8 0 ? 24 = 8 x 3 + 0
So the HCF of 56 and 32 is 8.
Therefore the fruit seller has to make s
REAL NUMBERS
S DIVISION ALGORITHM
Use Euclid’s division algorithm to find the HCF of 1362 and 6954
: Since 6954>1362 we apply division lemma to 6954 and 1362 to get,
6 9 5 4 1 3 6 2 x 5 1 4 4 = +
144 0 ? we apply the division lemma to 1362 and 144
1 3 6 2 1 4 4 x 9 + 6 6 =
Since the remainder 66 0 ? we apply the division lemma to 144 and 66
1 4 4 6 6 x 2 + 1 2 =
Since the remainder 12 0 ? we apply the division lemma to 66 and 12
6 6 1 2 x 5 + 6
Since the remainder 6 0 ? we apply the division lemma to 12 and 6
1 2 6 x 2 + 0
The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 6, the
HCF of 1362 and 6954 is 6.
odd integer is of the form 8q+1, or 8q+3, or 8q+5, or 8q+7
is a positive odd integer. We apply the division algorithm with
the possible remainders are 0,1,2,3,4,5,6 and 7
+2 or 8q+3 or 8q+4 or 8q+5 or 8q+6 or 8q+7 where
cannot be 8q or 8q+2 or 8q+4 or 8q+6 (since they are
integer can only be of the form 8 1 q+ , or 8 3 q+ , or 8 5 q+ , or
32 green apples. She wants to stack them in such a way
has the same number, and they take up the least area of the tray. What is the maximum number of
that can be placed in each stack for this purpose?
irst find the HCF of 56 and 32. Then this number will give us the
maximum number of apples in each stack and as a result the number of stacks will be least and they will
hm to find the HCF of 56 and 32
32 x 1 + 24
24 x 1 + 8
+ 0
Therefore the fruit seller has to make stacks of 8 for both kinds of apples.
: Since 6954>1362 we apply division lemma to 6954 and 1362 to get,
we apply the division lemma to 1362 and 144
e apply the division lemma to 144 and 66
we apply the division lemma to 66 and 12
we apply the division lemma to 12 and 6
The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 6, the
+7, where q is some integer
is a positive odd integer. We apply the division algorithm with a
+7 where q is the quotient.
(since they are divisible by 2)
, or 8 7 q+
in such a way that each stack
What is the maximum number of apples
Then this number will give us the
and as a result the number of stacks will be least and they will
Page 2
REAL
EUCLID’S DIVISION ALGORITHM
1) Use Euclid’s division algorithm to find the HCF of 1362 and 6954
Solution:
STEP 1: Since 6954>1362 we apply division lemma to 6954 and 1362 to get,
6 9 5 4 1 3 6 2 x 5 1 4 4 = +
STEP 2: Since the remainder 144 0
1 3 6 2 1 4 4 x 9 + 6 6
STEP 3: Since the remainder
1 4 4 6 6 x 2 + 1 2 =
STEP 4: Since the remainder
6 6 1 2 x 5 + 6 =
STEP 5: Since the remainder
1 2 6 x 2 + 0 =
The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 6, the
HCF of 1362 and 6954 is 6.
2) Show that any positive odd integer is of the form
Solution:
Let’s start with taking a, where a is a positive odd integer. We apply the division algorithm with
and b =8
Since 0 8 r = < the possible remainders are
That is, a can be 8q or 8q+1 or 8q+2 or 8
However, since a is odd a cannot be 8
Therefore any odd integer can only be of the form
3) A fruit seller has 56 red apples and
has the same number, and they take up the least area of the tray.
that can be placed in each stack for this purpose?
Solution:
To find the solution we have to first find the HCF of 56 and 32
maximum number of apples in each stack
take up least amount of space.
We use Euclid’s division algorithm to find the HCF of 56 and 32
56 = 32 x 1 + 24
Since, 24 0 ? ? 32 = 24 x 1 + 8
8 0 ? 24 = 8 x 3 + 0
So the HCF of 56 and 32 is 8.
Therefore the fruit seller has to make s
REAL NUMBERS
S DIVISION ALGORITHM
Use Euclid’s division algorithm to find the HCF of 1362 and 6954
: Since 6954>1362 we apply division lemma to 6954 and 1362 to get,
6 9 5 4 1 3 6 2 x 5 1 4 4 = +
144 0 ? we apply the division lemma to 1362 and 144
1 3 6 2 1 4 4 x 9 + 6 6 =
Since the remainder 66 0 ? we apply the division lemma to 144 and 66
1 4 4 6 6 x 2 + 1 2 =
Since the remainder 12 0 ? we apply the division lemma to 66 and 12
6 6 1 2 x 5 + 6
Since the remainder 6 0 ? we apply the division lemma to 12 and 6
1 2 6 x 2 + 0
The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 6, the
HCF of 1362 and 6954 is 6.
odd integer is of the form 8q+1, or 8q+3, or 8q+5, or 8q+7
is a positive odd integer. We apply the division algorithm with
the possible remainders are 0,1,2,3,4,5,6 and 7
+2 or 8q+3 or 8q+4 or 8q+5 or 8q+6 or 8q+7 where
cannot be 8q or 8q+2 or 8q+4 or 8q+6 (since they are
integer can only be of the form 8 1 q+ , or 8 3 q+ , or 8 5 q+ , or
32 green apples. She wants to stack them in such a way
has the same number, and they take up the least area of the tray. What is the maximum number of
that can be placed in each stack for this purpose?
irst find the HCF of 56 and 32. Then this number will give us the
maximum number of apples in each stack and as a result the number of stacks will be least and they will
hm to find the HCF of 56 and 32
32 x 1 + 24
24 x 1 + 8
+ 0
Therefore the fruit seller has to make stacks of 8 for both kinds of apples.
: Since 6954>1362 we apply division lemma to 6954 and 1362 to get,
we apply the division lemma to 1362 and 144
e apply the division lemma to 144 and 66
we apply the division lemma to 66 and 12
we apply the division lemma to 12 and 6
The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 6, the
+7, where q is some integer
is a positive odd integer. We apply the division algorithm with a
+7 where q is the quotient.
(since they are divisible by 2)
, or 8 7 q+
in such a way that each stack
What is the maximum number of apples
Then this number will give us the
and as a result the number of stacks will be least and they will
REAL NUMBERS
EUCLID’S DIVISION ALGORITHM
4) Prove that one of every four consecutive integers is divisible by 4
Solution:
Let the four consecutive integers be n,n
Consider a = bq + r, where 0= <
Let a=n and b=4. Hence r can have values 0
CASE 1: Let r=0 ? 4 n q =
CASE 2: Let r=1 ? 4 n q =
CASE 3: Let r=2 ? 4 n q =
CASE 4: Let r=3 ? 4 n q = +3 then
Thus in every case one of the four numbe
:
5) Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3
for some integer m.
Solution:
Let n be any positive integer. Consider
Let a=n and b=3. Hence r can have values 0,1 or 2.
CASE 1: Let r=0 then n =3q ? =
CASE 2: Let r=1 then 3 1 n q = +
CASE 2: Let r=2 then 3 2 n q = +
Therefore
2
n is always of the form
REAL NUMBERS
S DIVISION ALGORITHM
Prove that one of every four consecutive integers is divisible by 4.
Let the four consecutive integers be n,n+1,n+2 and n+3.
0 r b = <
can have values 0, 1, 2 or 3.
n q which is divisible by 4 :
n q +1 then 3 n+ = 4 1 3 4 4 4( 1) q q q + + = + = + which is divisible by 4
n q +2 then 2 n+ = 4 2 2 4 4 4( 1) q q q + + = + = +
n q +3 then 1 n+ = 4 3 1 4 4 4( 1) q q q + + = + = +
Thus in every case one of the four numbers is divisible by 4.
Euclid’s division lemma to show that the square of any positive integer is either of the form 3
Let n be any positive integer. Consider a=bq+r where 0 r b = <
can have values 0,1 or 2.
2 2
(3 ) n q ? =
2
9q =
2
3(3 ) q = 3m =
where m = 3q
3 1 n q = +
2 2
(3 1) n q ? = +
2
9 6 1 q q = + +
2
3(3 2 ) 1 q q = + +
3 1 m = +
where
2
3 2 m q q = +
3 2 = +
2 2
(3 2) n q ? = +
2
9 12 4 q q = + +
2 2
9 12 3 1 3(3 4 1) 1 q q q q = + + + = + + +
3 1 m = + where
is always of the form 3m or 3m+1
REAL NUMBERS
4 1 3 4 4 4( 1) which is divisible by 4 :
4 2 2 4 4 4( 1) + + = + = + which is divisible by 4 :
4 3 1 4 4 4( 1) which is divisible by 4
Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m+1
r b = <
m = 3q
2
9 12 3 1 3(3 4 1) 1 = + + + = + + +
where
2
3 4 1 m q q = + +
Page 3
REAL
EUCLID’S DIVISION ALGORITHM
1) Use Euclid’s division algorithm to find the HCF of 1362 and 6954
Solution:
STEP 1: Since 6954>1362 we apply division lemma to 6954 and 1362 to get,
6 9 5 4 1 3 6 2 x 5 1 4 4 = +
STEP 2: Since the remainder 144 0
1 3 6 2 1 4 4 x 9 + 6 6
STEP 3: Since the remainder
1 4 4 6 6 x 2 + 1 2 =
STEP 4: Since the remainder
6 6 1 2 x 5 + 6 =
STEP 5: Since the remainder
1 2 6 x 2 + 0 =
The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 6, the
HCF of 1362 and 6954 is 6.
2) Show that any positive odd integer is of the form
Solution:
Let’s start with taking a, where a is a positive odd integer. We apply the division algorithm with
and b =8
Since 0 8 r = < the possible remainders are
That is, a can be 8q or 8q+1 or 8q+2 or 8
However, since a is odd a cannot be 8
Therefore any odd integer can only be of the form
3) A fruit seller has 56 red apples and
has the same number, and they take up the least area of the tray.
that can be placed in each stack for this purpose?
Solution:
To find the solution we have to first find the HCF of 56 and 32
maximum number of apples in each stack
take up least amount of space.
We use Euclid’s division algorithm to find the HCF of 56 and 32
56 = 32 x 1 + 24
Since, 24 0 ? ? 32 = 24 x 1 + 8
8 0 ? 24 = 8 x 3 + 0
So the HCF of 56 and 32 is 8.
Therefore the fruit seller has to make s
REAL NUMBERS
S DIVISION ALGORITHM
Use Euclid’s division algorithm to find the HCF of 1362 and 6954
: Since 6954>1362 we apply division lemma to 6954 and 1362 to get,
6 9 5 4 1 3 6 2 x 5 1 4 4 = +
144 0 ? we apply the division lemma to 1362 and 144
1 3 6 2 1 4 4 x 9 + 6 6 =
Since the remainder 66 0 ? we apply the division lemma to 144 and 66
1 4 4 6 6 x 2 + 1 2 =
Since the remainder 12 0 ? we apply the division lemma to 66 and 12
6 6 1 2 x 5 + 6
Since the remainder 6 0 ? we apply the division lemma to 12 and 6
1 2 6 x 2 + 0
The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 6, the
HCF of 1362 and 6954 is 6.
odd integer is of the form 8q+1, or 8q+3, or 8q+5, or 8q+7
is a positive odd integer. We apply the division algorithm with
the possible remainders are 0,1,2,3,4,5,6 and 7
+2 or 8q+3 or 8q+4 or 8q+5 or 8q+6 or 8q+7 where
cannot be 8q or 8q+2 or 8q+4 or 8q+6 (since they are
integer can only be of the form 8 1 q+ , or 8 3 q+ , or 8 5 q+ , or
32 green apples. She wants to stack them in such a way
has the same number, and they take up the least area of the tray. What is the maximum number of
that can be placed in each stack for this purpose?
irst find the HCF of 56 and 32. Then this number will give us the
maximum number of apples in each stack and as a result the number of stacks will be least and they will
hm to find the HCF of 56 and 32
32 x 1 + 24
24 x 1 + 8
+ 0
Therefore the fruit seller has to make stacks of 8 for both kinds of apples.
: Since 6954>1362 we apply division lemma to 6954 and 1362 to get,
we apply the division lemma to 1362 and 144
e apply the division lemma to 144 and 66
we apply the division lemma to 66 and 12
we apply the division lemma to 12 and 6
The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 6, the
+7, where q is some integer
is a positive odd integer. We apply the division algorithm with a
+7 where q is the quotient.
(since they are divisible by 2)
, or 8 7 q+
in such a way that each stack
What is the maximum number of apples
Then this number will give us the
and as a result the number of stacks will be least and they will
REAL NUMBERS
EUCLID’S DIVISION ALGORITHM
4) Prove that one of every four consecutive integers is divisible by 4
Solution:
Let the four consecutive integers be n,n
Consider a = bq + r, where 0= <
Let a=n and b=4. Hence r can have values 0
CASE 1: Let r=0 ? 4 n q =
CASE 2: Let r=1 ? 4 n q =
CASE 3: Let r=2 ? 4 n q =
CASE 4: Let r=3 ? 4 n q = +3 then
Thus in every case one of the four numbe
:
5) Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3
for some integer m.
Solution:
Let n be any positive integer. Consider
Let a=n and b=3. Hence r can have values 0,1 or 2.
CASE 1: Let r=0 then n =3q ? =
CASE 2: Let r=1 then 3 1 n q = +
CASE 2: Let r=2 then 3 2 n q = +
Therefore
2
n is always of the form
REAL NUMBERS
S DIVISION ALGORITHM
Prove that one of every four consecutive integers is divisible by 4.
Let the four consecutive integers be n,n+1,n+2 and n+3.
0 r b = <
can have values 0, 1, 2 or 3.
n q which is divisible by 4 :
n q +1 then 3 n+ = 4 1 3 4 4 4( 1) q q q + + = + = + which is divisible by 4
n q +2 then 2 n+ = 4 2 2 4 4 4( 1) q q q + + = + = +
n q +3 then 1 n+ = 4 3 1 4 4 4( 1) q q q + + = + = +
Thus in every case one of the four numbers is divisible by 4.
Euclid’s division lemma to show that the square of any positive integer is either of the form 3
Let n be any positive integer. Consider a=bq+r where 0 r b = <
can have values 0,1 or 2.
2 2
(3 ) n q ? =
2
9q =
2
3(3 ) q = 3m =
where m = 3q
3 1 n q = +
2 2
(3 1) n q ? = +
2
9 6 1 q q = + +
2
3(3 2 ) 1 q q = + +
3 1 m = +
where
2
3 2 m q q = +
3 2 = +
2 2
(3 2) n q ? = +
2
9 12 4 q q = + +
2 2
9 12 3 1 3(3 4 1) 1 q q q q = + + + = + + +
3 1 m = + where
is always of the form 3m or 3m+1
REAL NUMBERS
4 1 3 4 4 4( 1) which is divisible by 4 :
4 2 2 4 4 4( 1) + + = + = + which is divisible by 4 :
4 3 1 4 4 4( 1) which is divisible by 4
Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m+1
r b = <
m = 3q
2
9 12 3 1 3(3 4 1) 1 = + + + = + + +
where
2
3 4 1 m q q = + +
REAL NUMBERS
THE FUNDAMENTAL THEOREM OF ARITHEMATICS
6) Find the HCF and LCM of the following pairs of numbers and verify that HCF
numbers.
a)39 and 169 b) 436 and 594
Solution: a) 39 and 169
Factors of the two numbers are
39
169
HCF(39,169) = 13
LCM(39,169) = 3 x 13 x 13=507
HCF x LCM = 13 x 507= 6591
Product of the numbers =39 x 169=6591
HCF x LCM= Product of the numbers
b)436 and 594
Factors of the two numbers are
436=2 x 2 x
594=2 x 3 x 3 x 3 x 11
HCF(436,594) = 2
LCM(436,594) = 2 x 2 x 3 x 3 x 3 x 11 x 109=129492
HCF x LCM = 2 x 129492=258984
Product of the two numbers = 436 x 594= 258984
? HCF x LCM= Product of the numbers
7) If HCF(90,225)=15, find LCM(90,225)
Solution: We know that HCF x LCM
? HCF(90,225) LCM(90,225) =
? 15 LCM(90,,225) = 90
? LCM
8) Find whether 8
n
can end with the digit 0 for any natural number.
Solution: The prime factors of 8 are 2.Therefore the prime factors of 8
Any digit ending with 0 must have 5 as one of its prime factors.
But the prime factors of 8 are 2.
Therefore 8
n
cannot end with 0.
REAL NUMBERS
THE FUNDAMENTAL THEOREM OF ARITHEMATICS
Find the HCF and LCM of the following pairs of numbers and verify that HCF x LCM=Product of the two
b) 436 and 594
are
39 = 3 x 13
169 = 13 x 13
3 x 13 x 13=507
= 13 x 507= 6591
Product of the numbers =39 x 169=6591
HCF x LCM= Product of the numbers
are
436=2 x 2 x 109
594=2 x 3 x 3 x 3 x 11
2 x 2 x 3 x 3 x 3 x 11 x 109=129492
= 2 x 129492=258984
Product of the two numbers = 436 x 594= 258984
HCF x LCM= Product of the numbers
LCM(90,225)
: We know that HCF x LCM = product of the two numbers
LCM(90,225) = product of (90,225)
LCM(90,,225) = 90 225
LCM(90,225) =
90 x 225
15
20250
15
= =1350
? LCM(90,225) = 1350
can end with the digit 0 for any natural number.
factors of 8 are 2.Therefore the prime factors of 8
n
will also be 2.
Any digit ending with 0 must have 5 as one of its prime factors.
But the prime factors of 8 are 2.
cannot end with 0.
LCM=Product of the two
will also be 2.
Page 4
REAL
EUCLID’S DIVISION ALGORITHM
1) Use Euclid’s division algorithm to find the HCF of 1362 and 6954
Solution:
STEP 1: Since 6954>1362 we apply division lemma to 6954 and 1362 to get,
6 9 5 4 1 3 6 2 x 5 1 4 4 = +
STEP 2: Since the remainder 144 0
1 3 6 2 1 4 4 x 9 + 6 6
STEP 3: Since the remainder
1 4 4 6 6 x 2 + 1 2 =
STEP 4: Since the remainder
6 6 1 2 x 5 + 6 =
STEP 5: Since the remainder
1 2 6 x 2 + 0 =
The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 6, the
HCF of 1362 and 6954 is 6.
2) Show that any positive odd integer is of the form
Solution:
Let’s start with taking a, where a is a positive odd integer. We apply the division algorithm with
and b =8
Since 0 8 r = < the possible remainders are
That is, a can be 8q or 8q+1 or 8q+2 or 8
However, since a is odd a cannot be 8
Therefore any odd integer can only be of the form
3) A fruit seller has 56 red apples and
has the same number, and they take up the least area of the tray.
that can be placed in each stack for this purpose?
Solution:
To find the solution we have to first find the HCF of 56 and 32
maximum number of apples in each stack
take up least amount of space.
We use Euclid’s division algorithm to find the HCF of 56 and 32
56 = 32 x 1 + 24
Since, 24 0 ? ? 32 = 24 x 1 + 8
8 0 ? 24 = 8 x 3 + 0
So the HCF of 56 and 32 is 8.
Therefore the fruit seller has to make s
REAL NUMBERS
S DIVISION ALGORITHM
Use Euclid’s division algorithm to find the HCF of 1362 and 6954
: Since 6954>1362 we apply division lemma to 6954 and 1362 to get,
6 9 5 4 1 3 6 2 x 5 1 4 4 = +
144 0 ? we apply the division lemma to 1362 and 144
1 3 6 2 1 4 4 x 9 + 6 6 =
Since the remainder 66 0 ? we apply the division lemma to 144 and 66
1 4 4 6 6 x 2 + 1 2 =
Since the remainder 12 0 ? we apply the division lemma to 66 and 12
6 6 1 2 x 5 + 6
Since the remainder 6 0 ? we apply the division lemma to 12 and 6
1 2 6 x 2 + 0
The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 6, the
HCF of 1362 and 6954 is 6.
odd integer is of the form 8q+1, or 8q+3, or 8q+5, or 8q+7
is a positive odd integer. We apply the division algorithm with
the possible remainders are 0,1,2,3,4,5,6 and 7
+2 or 8q+3 or 8q+4 or 8q+5 or 8q+6 or 8q+7 where
cannot be 8q or 8q+2 or 8q+4 or 8q+6 (since they are
integer can only be of the form 8 1 q+ , or 8 3 q+ , or 8 5 q+ , or
32 green apples. She wants to stack them in such a way
has the same number, and they take up the least area of the tray. What is the maximum number of
that can be placed in each stack for this purpose?
irst find the HCF of 56 and 32. Then this number will give us the
maximum number of apples in each stack and as a result the number of stacks will be least and they will
hm to find the HCF of 56 and 32
32 x 1 + 24
24 x 1 + 8
+ 0
Therefore the fruit seller has to make stacks of 8 for both kinds of apples.
: Since 6954>1362 we apply division lemma to 6954 and 1362 to get,
we apply the division lemma to 1362 and 144
e apply the division lemma to 144 and 66
we apply the division lemma to 66 and 12
we apply the division lemma to 12 and 6
The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 6, the
+7, where q is some integer
is a positive odd integer. We apply the division algorithm with a
+7 where q is the quotient.
(since they are divisible by 2)
, or 8 7 q+
in such a way that each stack
What is the maximum number of apples
Then this number will give us the
and as a result the number of stacks will be least and they will
REAL NUMBERS
EUCLID’S DIVISION ALGORITHM
4) Prove that one of every four consecutive integers is divisible by 4
Solution:
Let the four consecutive integers be n,n
Consider a = bq + r, where 0= <
Let a=n and b=4. Hence r can have values 0
CASE 1: Let r=0 ? 4 n q =
CASE 2: Let r=1 ? 4 n q =
CASE 3: Let r=2 ? 4 n q =
CASE 4: Let r=3 ? 4 n q = +3 then
Thus in every case one of the four numbe
:
5) Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3
for some integer m.
Solution:
Let n be any positive integer. Consider
Let a=n and b=3. Hence r can have values 0,1 or 2.
CASE 1: Let r=0 then n =3q ? =
CASE 2: Let r=1 then 3 1 n q = +
CASE 2: Let r=2 then 3 2 n q = +
Therefore
2
n is always of the form
REAL NUMBERS
S DIVISION ALGORITHM
Prove that one of every four consecutive integers is divisible by 4.
Let the four consecutive integers be n,n+1,n+2 and n+3.
0 r b = <
can have values 0, 1, 2 or 3.
n q which is divisible by 4 :
n q +1 then 3 n+ = 4 1 3 4 4 4( 1) q q q + + = + = + which is divisible by 4
n q +2 then 2 n+ = 4 2 2 4 4 4( 1) q q q + + = + = +
n q +3 then 1 n+ = 4 3 1 4 4 4( 1) q q q + + = + = +
Thus in every case one of the four numbers is divisible by 4.
Euclid’s division lemma to show that the square of any positive integer is either of the form 3
Let n be any positive integer. Consider a=bq+r where 0 r b = <
can have values 0,1 or 2.
2 2
(3 ) n q ? =
2
9q =
2
3(3 ) q = 3m =
where m = 3q
3 1 n q = +
2 2
(3 1) n q ? = +
2
9 6 1 q q = + +
2
3(3 2 ) 1 q q = + +
3 1 m = +
where
2
3 2 m q q = +
3 2 = +
2 2
(3 2) n q ? = +
2
9 12 4 q q = + +
2 2
9 12 3 1 3(3 4 1) 1 q q q q = + + + = + + +
3 1 m = + where
is always of the form 3m or 3m+1
REAL NUMBERS
4 1 3 4 4 4( 1) which is divisible by 4 :
4 2 2 4 4 4( 1) + + = + = + which is divisible by 4 :
4 3 1 4 4 4( 1) which is divisible by 4
Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m+1
r b = <
m = 3q
2
9 12 3 1 3(3 4 1) 1 = + + + = + + +
where
2
3 4 1 m q q = + +
REAL NUMBERS
THE FUNDAMENTAL THEOREM OF ARITHEMATICS
6) Find the HCF and LCM of the following pairs of numbers and verify that HCF
numbers.
a)39 and 169 b) 436 and 594
Solution: a) 39 and 169
Factors of the two numbers are
39
169
HCF(39,169) = 13
LCM(39,169) = 3 x 13 x 13=507
HCF x LCM = 13 x 507= 6591
Product of the numbers =39 x 169=6591
HCF x LCM= Product of the numbers
b)436 and 594
Factors of the two numbers are
436=2 x 2 x
594=2 x 3 x 3 x 3 x 11
HCF(436,594) = 2
LCM(436,594) = 2 x 2 x 3 x 3 x 3 x 11 x 109=129492
HCF x LCM = 2 x 129492=258984
Product of the two numbers = 436 x 594= 258984
? HCF x LCM= Product of the numbers
7) If HCF(90,225)=15, find LCM(90,225)
Solution: We know that HCF x LCM
? HCF(90,225) LCM(90,225) =
? 15 LCM(90,,225) = 90
? LCM
8) Find whether 8
n
can end with the digit 0 for any natural number.
Solution: The prime factors of 8 are 2.Therefore the prime factors of 8
Any digit ending with 0 must have 5 as one of its prime factors.
But the prime factors of 8 are 2.
Therefore 8
n
cannot end with 0.
REAL NUMBERS
THE FUNDAMENTAL THEOREM OF ARITHEMATICS
Find the HCF and LCM of the following pairs of numbers and verify that HCF x LCM=Product of the two
b) 436 and 594
are
39 = 3 x 13
169 = 13 x 13
3 x 13 x 13=507
= 13 x 507= 6591
Product of the numbers =39 x 169=6591
HCF x LCM= Product of the numbers
are
436=2 x 2 x 109
594=2 x 3 x 3 x 3 x 11
2 x 2 x 3 x 3 x 3 x 11 x 109=129492
= 2 x 129492=258984
Product of the two numbers = 436 x 594= 258984
HCF x LCM= Product of the numbers
LCM(90,225)
: We know that HCF x LCM = product of the two numbers
LCM(90,225) = product of (90,225)
LCM(90,,225) = 90 225
LCM(90,225) =
90 x 225
15
20250
15
= =1350
? LCM(90,225) = 1350
can end with the digit 0 for any natural number.
factors of 8 are 2.Therefore the prime factors of 8
n
will also be 2.
Any digit ending with 0 must have 5 as one of its prime factors.
But the prime factors of 8 are 2.
cannot end with 0.
LCM=Product of the two
will also be 2.
REAL NUMBERS
THE FUNDAMENTAL THEOREM OF ARITHEMATICS
9) Find the HCF and LCM of the following numbers using prime factorisation method
a) 9, 27 and 108 b) 16, 56 and 144
Solution:
a) 9, 27 and 108
Factors of the numbers are
9 = 3 x 3
27 = 3 x 3 x 3
108 = 2 x 2 x 3 x 3 x 3
HCF(9,27,108)=3 x 3=9
LCM(9,27,108)=2 x 2 x 3 x 3 x 3= 108
? HCF=9 & LCM=108
b) 16, 56 and 144
Factors of the numbers are
16 = 2 x 2 x 2 x 2
56 = 2 x 2 x 2 x 7
144 = 2 x 2 x 2 x 2 x 3
?HCF (16,56,144)=2 x 2 x 2=8
?LCM (16,56,144)=2 x 2 x 2 x 2 x 3 x 3 x 7=1008
?
HCF = 8 & LCM = 1008
10) Prove that 16, 343 and 225 are relatively
Solution:
Factors of the numbers
16 =2 x 2 x 2 x 2
343 = 7 x 7 x 7
225 = 3 x 3 x 5 x 5
?HCF(16,343,225)=1
That is they do not have any common factors
? They are relatively prime.
REAL NUMBERS
THE FUNDAMENTAL THEOREM OF ARITHEMATICS
Find the HCF and LCM of the following numbers using prime factorisation method
108 b) 16, 56 and 144
3 x 3
3 x 3 x 3
2 x 2 x 3 x 3 x 3
LCM(9,27,108)=2 x 2 x 3 x 3 x 3= 108
2 x 2 x 2 x 2
2 x 2 x 2 x 7
2 x 2 x 2 x 2 x 3 x 3
HCF (16,56,144)=2 x 2 x 2=8
(16,56,144)=2 x 2 x 2 x 2 x 3 x 3 x 7=1008
and 225 are relatively prime.
=2 x 2 x 2 x 2
= 7 x 7 x 7
= 3 x 3 x 5 x 5
That is they do not have any common factors
REAL NUMBERS
Find the HCF and LCM of the following numbers using prime factorisation method
Page 5
REAL
EUCLID’S DIVISION ALGORITHM
1) Use Euclid’s division algorithm to find the HCF of 1362 and 6954
Solution:
STEP 1: Since 6954>1362 we apply division lemma to 6954 and 1362 to get,
6 9 5 4 1 3 6 2 x 5 1 4 4 = +
STEP 2: Since the remainder 144 0
1 3 6 2 1 4 4 x 9 + 6 6
STEP 3: Since the remainder
1 4 4 6 6 x 2 + 1 2 =
STEP 4: Since the remainder
6 6 1 2 x 5 + 6 =
STEP 5: Since the remainder
1 2 6 x 2 + 0 =
The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 6, the
HCF of 1362 and 6954 is 6.
2) Show that any positive odd integer is of the form
Solution:
Let’s start with taking a, where a is a positive odd integer. We apply the division algorithm with
and b =8
Since 0 8 r = < the possible remainders are
That is, a can be 8q or 8q+1 or 8q+2 or 8
However, since a is odd a cannot be 8
Therefore any odd integer can only be of the form
3) A fruit seller has 56 red apples and
has the same number, and they take up the least area of the tray.
that can be placed in each stack for this purpose?
Solution:
To find the solution we have to first find the HCF of 56 and 32
maximum number of apples in each stack
take up least amount of space.
We use Euclid’s division algorithm to find the HCF of 56 and 32
56 = 32 x 1 + 24
Since, 24 0 ? ? 32 = 24 x 1 + 8
8 0 ? 24 = 8 x 3 + 0
So the HCF of 56 and 32 is 8.
Therefore the fruit seller has to make s
REAL NUMBERS
S DIVISION ALGORITHM
Use Euclid’s division algorithm to find the HCF of 1362 and 6954
: Since 6954>1362 we apply division lemma to 6954 and 1362 to get,
6 9 5 4 1 3 6 2 x 5 1 4 4 = +
144 0 ? we apply the division lemma to 1362 and 144
1 3 6 2 1 4 4 x 9 + 6 6 =
Since the remainder 66 0 ? we apply the division lemma to 144 and 66
1 4 4 6 6 x 2 + 1 2 =
Since the remainder 12 0 ? we apply the division lemma to 66 and 12
6 6 1 2 x 5 + 6
Since the remainder 6 0 ? we apply the division lemma to 12 and 6
1 2 6 x 2 + 0
The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 6, the
HCF of 1362 and 6954 is 6.
odd integer is of the form 8q+1, or 8q+3, or 8q+5, or 8q+7
is a positive odd integer. We apply the division algorithm with
the possible remainders are 0,1,2,3,4,5,6 and 7
+2 or 8q+3 or 8q+4 or 8q+5 or 8q+6 or 8q+7 where
cannot be 8q or 8q+2 or 8q+4 or 8q+6 (since they are
integer can only be of the form 8 1 q+ , or 8 3 q+ , or 8 5 q+ , or
32 green apples. She wants to stack them in such a way
has the same number, and they take up the least area of the tray. What is the maximum number of
that can be placed in each stack for this purpose?
irst find the HCF of 56 and 32. Then this number will give us the
maximum number of apples in each stack and as a result the number of stacks will be least and they will
hm to find the HCF of 56 and 32
32 x 1 + 24
24 x 1 + 8
+ 0
Therefore the fruit seller has to make stacks of 8 for both kinds of apples.
: Since 6954>1362 we apply division lemma to 6954 and 1362 to get,
we apply the division lemma to 1362 and 144
e apply the division lemma to 144 and 66
we apply the division lemma to 66 and 12
we apply the division lemma to 12 and 6
The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 6, the
+7, where q is some integer
is a positive odd integer. We apply the division algorithm with a
+7 where q is the quotient.
(since they are divisible by 2)
, or 8 7 q+
in such a way that each stack
What is the maximum number of apples
Then this number will give us the
and as a result the number of stacks will be least and they will
REAL NUMBERS
EUCLID’S DIVISION ALGORITHM
4) Prove that one of every four consecutive integers is divisible by 4
Solution:
Let the four consecutive integers be n,n
Consider a = bq + r, where 0= <
Let a=n and b=4. Hence r can have values 0
CASE 1: Let r=0 ? 4 n q =
CASE 2: Let r=1 ? 4 n q =
CASE 3: Let r=2 ? 4 n q =
CASE 4: Let r=3 ? 4 n q = +3 then
Thus in every case one of the four numbe
:
5) Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3
for some integer m.
Solution:
Let n be any positive integer. Consider
Let a=n and b=3. Hence r can have values 0,1 or 2.
CASE 1: Let r=0 then n =3q ? =
CASE 2: Let r=1 then 3 1 n q = +
CASE 2: Let r=2 then 3 2 n q = +
Therefore
2
n is always of the form
REAL NUMBERS
S DIVISION ALGORITHM
Prove that one of every four consecutive integers is divisible by 4.
Let the four consecutive integers be n,n+1,n+2 and n+3.
0 r b = <
can have values 0, 1, 2 or 3.
n q which is divisible by 4 :
n q +1 then 3 n+ = 4 1 3 4 4 4( 1) q q q + + = + = + which is divisible by 4
n q +2 then 2 n+ = 4 2 2 4 4 4( 1) q q q + + = + = +
n q +3 then 1 n+ = 4 3 1 4 4 4( 1) q q q + + = + = +
Thus in every case one of the four numbers is divisible by 4.
Euclid’s division lemma to show that the square of any positive integer is either of the form 3
Let n be any positive integer. Consider a=bq+r where 0 r b = <
can have values 0,1 or 2.
2 2
(3 ) n q ? =
2
9q =
2
3(3 ) q = 3m =
where m = 3q
3 1 n q = +
2 2
(3 1) n q ? = +
2
9 6 1 q q = + +
2
3(3 2 ) 1 q q = + +
3 1 m = +
where
2
3 2 m q q = +
3 2 = +
2 2
(3 2) n q ? = +
2
9 12 4 q q = + +
2 2
9 12 3 1 3(3 4 1) 1 q q q q = + + + = + + +
3 1 m = + where
is always of the form 3m or 3m+1
REAL NUMBERS
4 1 3 4 4 4( 1) which is divisible by 4 :
4 2 2 4 4 4( 1) + + = + = + which is divisible by 4 :
4 3 1 4 4 4( 1) which is divisible by 4
Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m+1
r b = <
m = 3q
2
9 12 3 1 3(3 4 1) 1 = + + + = + + +
where
2
3 4 1 m q q = + +
REAL NUMBERS
THE FUNDAMENTAL THEOREM OF ARITHEMATICS
6) Find the HCF and LCM of the following pairs of numbers and verify that HCF
numbers.
a)39 and 169 b) 436 and 594
Solution: a) 39 and 169
Factors of the two numbers are
39
169
HCF(39,169) = 13
LCM(39,169) = 3 x 13 x 13=507
HCF x LCM = 13 x 507= 6591
Product of the numbers =39 x 169=6591
HCF x LCM= Product of the numbers
b)436 and 594
Factors of the two numbers are
436=2 x 2 x
594=2 x 3 x 3 x 3 x 11
HCF(436,594) = 2
LCM(436,594) = 2 x 2 x 3 x 3 x 3 x 11 x 109=129492
HCF x LCM = 2 x 129492=258984
Product of the two numbers = 436 x 594= 258984
? HCF x LCM= Product of the numbers
7) If HCF(90,225)=15, find LCM(90,225)
Solution: We know that HCF x LCM
? HCF(90,225) LCM(90,225) =
? 15 LCM(90,,225) = 90
? LCM
8) Find whether 8
n
can end with the digit 0 for any natural number.
Solution: The prime factors of 8 are 2.Therefore the prime factors of 8
Any digit ending with 0 must have 5 as one of its prime factors.
But the prime factors of 8 are 2.
Therefore 8
n
cannot end with 0.
REAL NUMBERS
THE FUNDAMENTAL THEOREM OF ARITHEMATICS
Find the HCF and LCM of the following pairs of numbers and verify that HCF x LCM=Product of the two
b) 436 and 594
are
39 = 3 x 13
169 = 13 x 13
3 x 13 x 13=507
= 13 x 507= 6591
Product of the numbers =39 x 169=6591
HCF x LCM= Product of the numbers
are
436=2 x 2 x 109
594=2 x 3 x 3 x 3 x 11
2 x 2 x 3 x 3 x 3 x 11 x 109=129492
= 2 x 129492=258984
Product of the two numbers = 436 x 594= 258984
HCF x LCM= Product of the numbers
LCM(90,225)
: We know that HCF x LCM = product of the two numbers
LCM(90,225) = product of (90,225)
LCM(90,,225) = 90 225
LCM(90,225) =
90 x 225
15
20250
15
= =1350
? LCM(90,225) = 1350
can end with the digit 0 for any natural number.
factors of 8 are 2.Therefore the prime factors of 8
n
will also be 2.
Any digit ending with 0 must have 5 as one of its prime factors.
But the prime factors of 8 are 2.
cannot end with 0.
LCM=Product of the two
will also be 2.
REAL NUMBERS
THE FUNDAMENTAL THEOREM OF ARITHEMATICS
9) Find the HCF and LCM of the following numbers using prime factorisation method
a) 9, 27 and 108 b) 16, 56 and 144
Solution:
a) 9, 27 and 108
Factors of the numbers are
9 = 3 x 3
27 = 3 x 3 x 3
108 = 2 x 2 x 3 x 3 x 3
HCF(9,27,108)=3 x 3=9
LCM(9,27,108)=2 x 2 x 3 x 3 x 3= 108
? HCF=9 & LCM=108
b) 16, 56 and 144
Factors of the numbers are
16 = 2 x 2 x 2 x 2
56 = 2 x 2 x 2 x 7
144 = 2 x 2 x 2 x 2 x 3
?HCF (16,56,144)=2 x 2 x 2=8
?LCM (16,56,144)=2 x 2 x 2 x 2 x 3 x 3 x 7=1008
?
HCF = 8 & LCM = 1008
10) Prove that 16, 343 and 225 are relatively
Solution:
Factors of the numbers
16 =2 x 2 x 2 x 2
343 = 7 x 7 x 7
225 = 3 x 3 x 5 x 5
?HCF(16,343,225)=1
That is they do not have any common factors
? They are relatively prime.
REAL NUMBERS
THE FUNDAMENTAL THEOREM OF ARITHEMATICS
Find the HCF and LCM of the following numbers using prime factorisation method
108 b) 16, 56 and 144
3 x 3
3 x 3 x 3
2 x 2 x 3 x 3 x 3
LCM(9,27,108)=2 x 2 x 3 x 3 x 3= 108
2 x 2 x 2 x 2
2 x 2 x 2 x 7
2 x 2 x 2 x 2 x 3 x 3
HCF (16,56,144)=2 x 2 x 2=8
(16,56,144)=2 x 2 x 2 x 2 x 3 x 3 x 7=1008
and 225 are relatively prime.
=2 x 2 x 2 x 2
= 7 x 7 x 7
= 3 x 3 x 5 x 5
That is they do not have any common factors
REAL NUMBERS
Find the HCF and LCM of the following numbers using prime factorisation method
REAL NUMBERS
IRRATIONAL NUMBERS
11) Prove that 7 is irrational.
Solution: Let 7 be rational. ? There exists integers a, b such that
are co-prime.
7b a ? =
Squaring both sides we get,
2 2
7b a = Therefore
2
a is divisible by 7. Hence a is also divisible by 7.
? There exists an integer c such that
2 2
7 49 b c = and hence
2 2
b c
Therefore a and b have a common factor 7. This is a contradiction, since a and b are co
Hence our assumption 7
Therefore 7 is irrational
12) Prove that 6 - 5 is irrational
Solution: Let us assume that 6 5 -
that 6 5
a
b
- = where a and b are co
Rearranging the equation we get
Since a and b are integers
But this contradicts the fact that
Therefore our assumption that
Therefore 6 5 is irrational -
13) Prove that
3
4 is irrational.
Solution: Let us assume that
3
4 is rational. Therefore there exists integers a and b(
where a and b are co-prime
i.e.
3 3
4 a b = . Therefore
3
a
Therefore there exists an integer c such that
3 3
16 b c ? = Hence
Therefore a and b have a common factor 4
3
Therefore 4 is irrational
REAL NUMBERS
IRRATIONAL NUMBERS
There exists integers a, b such that 7
a
b
= where a and b
Squaring both sides we get,
is divisible by 7. Hence a is also divisible by 7.
integer c such that 7 a c =
2 2
7 (7 ) b c ? =
2 2
7 b c = which means
2
b is divisible by 7 and b is also divisible by 7
a and b have a common factor 7. This is a contradiction, since a and b are co
7 is rational is wrong.
Therefore 7 is irrational
6 5 is rational. Therefore there exists integers a and b
where a and b are co-prime.
Rearranging the equation we get 6 5
a
b
- =
also b0, 6
a
b
- is rational. Therefore 5 is rational
But this contradicts the fact that 5 is irrational.
Therefore our assumption that 6 5 - is rational is wrong.
Therefore 6 5 is irrational
is rational. Therefore there exists integers a and b(
prime Taking cubes on both sides we get,t
3
4
a
b
=
3
a is divisible by 4 Hence a is also divisible by 4
Therefore there exists an integer c such that a=4c.
3 3 3
4 (4 ) 64 b c c ? = =
Hence
3
b is divisible by 4 and hence b is divisible by 4.
e a and b have a common factor 4. This is a contradiction, since a and b are co
Therefore 4 is irrational
where a and b( 0 b? )
is divisible by 7. Hence a is also divisible by 7.
is divisible by 7 and b is also divisible by 7
a and b have a common factor 7. This is a contradiction, since a and b are co-prime.
is rational. Therefore there exists integers a and b( 0 b? ) such
is rational
is rational. Therefore there exists integers a and b( 0 b? ) such that
3
4
a
b
=
3
3
a
b
divisible by 4
3 3 3
b c c
and hence b is divisible by 4.
. This is a contradiction, since a and b are co-prime
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