EXERCISE 1.1
1. Use Euclid's division algorithm to find the HCF of:
@ 135 and 225 (m) 196 and 38220
(m) 867 and 255
2. Show that any positive odd integer is of the form 6q+1. or 60 + 3. or 6q+5, where is
some integer (lomon)
3. An army contingent of 616 members is to march behind an army band of 32 members in
a parade. The two groups are to march in the same number of columns. What is the
maximum number of columns in which they can march?
4. Use Euclid's division lemma to show that the square of any positive integer is either of
the form 3m or 3m+T for some integer m.
Hint: Letx be any positive integer then it is of the form 39,39 +1 or 39+2. Now square
each of these and show that they can be rewritten in the form 3m or 3m+ 1.)
5 e Euclid's division lemma to show that the cube of any positive integer is of the form
9m, 9m+ 1 or 9m+&
13 The Fundamental Theorem of Arithmetic
less you have seen that any natural
Answers
Answer:
Step-by-step explanation:
1. (i) 225 = 135 x 1 + 120
135 = 120 x 1 + 15
120 = 15 x 8 +0
∴ HCF ( 225,135 ) is 15.
(ii) 38220 = 196 x 195 + 0
∴ HCF ( 38220,196 ) is 195.
(iii) 867 = 225 x 3 + 102
225 = 102 x 2 + 51
102 = 51 x 2 + 0
∴HCF ( 867,225 ) is 51.
2. Any odd positive integer is always of the form (2m + 1) → 1
Let a be the positive odd integer which when divided by 6 gives q as quotient and r as remainder.
according to Euclid's division lemma
a = bq + r
⇒ a = 6q + r
where r = 0 , 1 , 2 , 3 , 4 , 5
∴When r = 0,
a = 6q + 0
= 2(3q)
⇒it is an even no as it is of the form 2m where m = 3q.
When r = 1
a = 6q + 1
= 2(3q) + 1
⇒it is an odd no as it is of the form 2m + 1 where m = 3q.
When r = 2
a = 6q + 2
= 2(3q + 1)
⇒it is an even no as it is of the form 2m where m = 3q + 1
When r = 3
a = 6q + 3
a = 6q + 2 + 1
= 2(3q + 1) +1
⇒it is an odd no as it is of the form 2m + 1 where m = 3q+1
When r = 4
a = 6q + 4
= 2(3q + 2)
⇒it is an even no as it is of the form 2m where m = 3q + 1
When r = 5
a = 6q + 5
a = 6q + 4 + 1
= 2(3q + 2) +1
⇒it is an odd no as it is of the form 2m + 1 where m = 3q+2
∴ Hence proved.
3.To get the maximum number column here we always find HCF and for minimum number we find LCM
So can use Euclid’s algorithm to find the HCF.
Here 616> 32 so always divide greater number with smaller one
When we divide 616 by 32 we get quotient 19 and remainder 8
So we can write it as
616 = 32 x 19 + 8
32 = 8 x 4 + 0
As there are no remainder so our HCF will 8.
4. Let us take α as any positive integer and b = 3.
Then using Euclid’s algorithm we get a = 3q + r here r is remainder and value of q is more than or equal to 0 and r = 0, 1, 2 because 0 < r < b and the value of b is 3 So our possible values will 3q+0 , 3q+1 and 3q+2
Now find the square of values
Use the formula (a+b)² = a² + 2ab +b² to open the square bracket
(3q)² = 9q²
if we divide by 3 we get no remainder
we can write it as 3*(3q²) so it is in form of 3m here m = 3q²
(3q+1)² = (3q)² + 2 x 3q x 1 + 1²
=9q² + 6q +1 now divide by 3 we get 1 remainder
so we can write it as 3(3q² + 2q) +1 so we can write it in form of 3m+1 and value of m is 3q² + 2q here
(3q+2)² = (3q)² + 2*3q*2 + 2²
=9q² + 12q + 4
now divide by 3 we get 1 remainder
so we can write it as 3(3q² + 4q +1) +1 so we can write it in form of 3m +1 and value of m will 3q² + 4q +1
Square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
∴ Hence proved.
5.According to Euclid’s Division Lemma
Let take a as any positive integer and b = 9.
Then using Euclid’s algorithm we get a = 9q + r here r is remainder and value of q is more than or equal to 0 and r = 0, 1, 2, 3, 4, 5 , 6 , 7 , 8, because 0 ≤r < b and the value of b is 9
Sp possible forms will 9q, 9q+1, 9q+2,9q+3,9q+4,9q+5,9q+6,9q+7 and 9q+8
to get the cube of these values use the formula
(a+b)³ = a³ + 3a²b+ 3ab² + b³
In this formula value of a is always 9q
So plug the value we get
(9q+b)³ = 729q³ + 243q²b + 27qb² + b³
Now divide by 9 we get quotient = 81q³ + 27q²b + 3qb² and remainder is b³
So we have to consider the value of b³
b = 0
we get 9m+0 = 9m
b = 1
then 1³ = 1 so we get 9m +1
b = 2
then 2³ = 8 so we get 9m + 8
b = 3
then 3³ = 27 and it is divisible by 9 so we get 9m
b = 4
then 4³ =64 divide by 9 we get 1 as remainder so we get 9m +1
b=6
then 6³=216 divide by 9 no remainder there so we get 9m
b=7
then 7³ = 343 divide by 9 we get 1 as remainder so we get 9m+1
b=8
then 8³ = 512 divide by 9 we get 8 as remainder so we get 9m+8
∴ All values are in form of 9m , 9m+1 or 9m+8 .
13.The fundamental theorem of arithmetic states that every positive integer (except the number 1) can be represented in exactly one way apart from rearrangement as a product of one or more primes This theorem is also called the unique factorization theorem.