## NCERT EXERCISE - 1.1

### Question - 1

#### Use Euclid’s division algorithm to find the HCF of : (i) 135 and 225(ii) 196 and 38220(iii) 867 and 255Solution:

(i) Finding HCF(135,225)
Since, 225>135,
Applying Euclid’s Lemma to 225 and 135, we get,
225 = 135 $$\times$$ 1 $$+$$ 90
Since, $$Remainder \ne 0$$,
Applying Euclid’s Lemma to 135 and 90, we get,
135 = 90 $$\times$$ 1 $$+$$ 45
Since, $$Remainder \ne 0$$,
Applying Euclid’s Lemma to 90 and 45, we get,
90 = 45 $$\times$$ 2 $$+$$ 0
Since, $$Remainder = 0$$,
HCF(225,135) = 45

(ii) Finding HCF(196,38220)
Since, 38220 > 196,
Applying Euclid’s Lemma to 38220 and 196, we get,
38220 = 196 $$\times$$ 195 $$+$$ 0
Since, $$Remainder = 0$$,
HCF(38220,196) = 196

(iii) Finding HCF(867,255)
Since, 867>255,
Applying Euclid’s Lemma to 867 and 255, we get,
867 = 255 $$\times$$ 3 $$+$$ 102
Since, $$Remainder \ne 0$$,
Applying Euclid’s Lemma to 255 and 102, we get,
255 = 102 $$\times$$ 2 $$+$$ 51
Since, $$Remainder \ne 0$$,
Applying Euclid’s Lemma to 102 and 51, we get,
102 = 51 $$\times$$ 2 $$+$$ 0
Since, $$Remainder = 0$$,
HCF(867,255) = 51

### Question - 2

#### Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.Solution:

Let a be any positive integer and b=6, where $$a>b$$.
Applying Euclid’s Lemma on a and b, we get,
$$a = 6\times q + r$$, where $$0\le r<6$$.
If $$r=0$$
$$a = 6q$$, which is an even number.
If $$r = 1$$
$$a = 6q + 1$$, which is an odd number.
If $$r = 2$$
$$a = 6q + 2 = 2(3q + 1)$$, which is an even number.
If $$r = 3$$
$$a = 6q + 3 = 3(2q + 1)$$, which is an odd number.
If $$r = 4$$
$$a = 6q + 4 = 2(3q + 2)$$, which is an even number.
If $$r = 5$$
$$a = 6q + 5$$, which is an odd number.

Hence, every positive odd integer is of the form $$6q + 1$$, or $$6q + 3$$, or $$6q + 5$$.

### Question - 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?Solution:

Finding HCF(616,32)
Since, 616>32,
Applying Euclid’s Lemma to 616 and 32, we get,
616 = 32 $$\times$$ 19 $$+$$ 8
Since, $$Remainder \ne 0$$,
Applying Euclid’s Lemma to 32 and 8, we get,
32 = 8 $$\times$$ 4 $$+$$ 0
Since, $$Remainder = 0$$,
HCF(616,32) = 8

### Question - 4

#### Use Euclid’s division lemma to show that the square of any positive integer is either of the form $$3m$$ or $$3m + 1$$ for some integer $$m$$.Solution:

Let a be any positive integer and b=3, where $$a>b$$.
Applying Euclid’s Lemma on a and b, we get,
$$a = 3\times q + r$$, where $$0\le r<3$$.

If $$r=0$$
$$a = 3q$$
Squaring both sides, we get,
$$a^2=9q^2=3(3q^2) = 3m$$, where $$m =3q^2$$

If $$r = 1$$
$$a = 3q+1$$
Squaring both sides, we get,
$$a^2=9q^2+6q+1$$,
$$\implies a^2=3(3q^2+2q)+1$$
$$\implies a^2=3m+1$$, where $$m =(3q^2+2q)$$

If $$r = 2$$
$$a = 3q+2$$
Squaring both sides, we get,
$$a^2=9q^2+12q+4$$,
$$\implies a^2=(9q^2+12q+3)+1$$
$$\implies a^2=3(3q^2+4q+1)+1$$ $$\implies a^2=3m+1$$, where $$m =(3q^2+4q+1)$$

Hence, square of any positive integer is either of the form $$3m$$ or $$3m + 1$$ for some integer $$m$$.

### Question - 5

#### Use Euclid’s division lemma to show that the cube of any positive integer is of the form $$9m$$, $$9m + 1$$ or $$9m + 8$$.Solution:

Let a be any positive integer and b=3, where $$a>b$$.
Applying Euclid’s Lemma on a and b, we get,
$$a = 3\times q + r$$, where $$0\le r<3$$.

If $$r=0$$
$$a = 3q$$
Taking cube of both sides, we get,
$$a^3=27q^3=9(3q^3) = 3m$$, where $$m =3q^3$$

If $$r = 1$$
$$a = 3q+1$$
Taking cube of both sides, we get,
$$a^3=27q^3+3(3q)(1)(3q+1)+1$$,
$$\implies a^3=27q^3+27q^2+9q+1$$
$$\implies a^3=9(3q^3+3q^2+q)+1$$
$$\implies a^3=9m+1$$
, where $$m =(3q^3+3q^2+q)$$

If $$r = 2$$
$$a = 3q+2$$
Taking cube of both sides, we get,
$$a^3=27q^3+3(3q)(2)(3q+2)+8$$,
$$\implies a^3=27q^3+54q^2+36q+8$$
$$\implies a^3=9(3q^3+6q^2+4q)+8$$
$$\implies a^3=9m+8$$
, where $$m =(3q^3+6q^2+4q)$$

Hence, cube of any positive integer is of the form $$9m$$, $$9m + 1$$ or $$9m + 8$$.

## NCERT EXERCISE - 1.2

### Question - 1

#### Express each number as a product of its prime factors: (i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429Solution:

(i) 140
We know, $$\therefore$$ Product of prime factors of 140 = $$2\times 2\times 5\times 7 = 2^2\times 5 \times 7$$

(ii) 156
We know, $$\therefore$$ Product of prime factors of 156 = $$2\times 2\times 3\times 13 = 2^2\times 3 \times 13$$

(iii) 3825
We know, $$\therefore$$ Product of prime factors of 3825 = $$5\times 5\times 3\times 3\times 17 = 5^2\times 3^2 \times 17$$

(iv) 5005
We know, $$\therefore$$ Product of prime factors of 5005 = $$5\times 7\times 11\times 13$$

(v) 7429
We know, $$\therefore$$ Product of prime factors of 7429 = $$17\times 19\times 23$$

### Question - 2

#### Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers. (i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54Solution:

(i) 26 and 91
We know,
$$26 = 2\times13$$
$$91 = 7\times 13$$
$$\therefore$$ LCM (26,91) = $$2\times 7\times 13 = 182$$ and
HCF (26,91) = 13

Verifying: LCM $$\times$$ HCF = Product of two numbers
LHS: LCM $$\times$$ HCF = $$182\times13$$ = 2366
RHS: Product of two numbers = $$26\times 91$$ = 2366
$$\implies$$ LHS = RHS
Hence verified.

(ii) 510 and 92
We know,
$$510 = 2\times 3\times5 \times 17$$
$$92 = 2\times 2\times 23$$
$$\therefore$$ LCM (510,92) = $$2\times 2\times 3\times 5\times 17\times 23 = 23460$$ and
HCF (510,92) = 2

Verifying: LCM $$\times$$ HCF = Product of two numbers
LHS: LCM $$\times$$ HCF = $$23460\times 2$$ = 46920
RHS: Product of two numbers = $$510\times 92$$ = 46920
$$\implies$$ LHS = RHS
Hence verified.

(iii) 336 and 54
We know,
$$336 = 2\times 2\times 2\times 2\times 3\times 7 = 2^4\times 3\times 7$$
$$54 = 2\times 3\times 3\times 3= 2\times3^3$$
$$\therefore$$ LCM (336,54) = $$2^4\times 3^3 \times 7= 3024$$ and
HCF (336,54) = $$2\times 3 = 6$$

Verifying: LCM $$\times$$ HCF = Product of two numbers
LHS: LCM $$\times$$ HCF = $$3024\times 6$$ = 18144
RHS: Product of two numbers = $$336\times 54$$ = 18144 $$\implies$$ LHS = RHS
Hence verified.

### Question - 3

#### Find the LCM and HCF of the following integers by applying the prime factorisation method. (i) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25Solution:

(i) 12, 15 and 21
Writing the product of prime factors for all the three numbers, we get,
12 = 2 × 2 × 3
15 = 5 × 3
21 = 7 × 3
Therefore,
HCF(12,15,21) = 3
LCM(12,15,21) = 2 × 2 × 3 × 5 × 7 = 420

(ii) 17, 23 and 29
Writing the product of prime factors for all the three numbers, we get,
17 = 17 × 1
23 = 23 × 1
29 = 29 × 1
Therefore,
HCF(17,23,29) = 1
LCM(17,23,29) = 17 × 23 × 29 = 11339

(iii) 8, 9 and 25
Writing the product of prime factors for all the three numbers, we get,
8 = 2 × 2 × 2
9 = 3 × 3
25 = 5 × 5
Therefore,
HCF(8,9,25)=1
LCM(8,9,25) = 2 × 2 × 2 × 3 × 3 × 5 × 5 = 1800

### Question - 4

#### Given that HCF (306, 657) = 9, find LCM (306, 657).Solution:

Given : HCF = 9

We know that,
LCM $$\times$$ HCF = Product of two numbers

LCM $$\times$$ 9 = 306 $$\times$$ 657
$$LCM = \cfrac{306\times 657}{9}$$
$$\therefore$$ LCM = 22338

### Question - 5

#### Check whether $$6^n$$ can end with the digit 0 for any natural number n. Solution:

$$6^n$$ can end with digit 0 only if it is divisible by 5.
Prime Factorization of $$6^n$$ = $$(2 \times 3)^n$$
This is not divisible by 5 because $$6^n$$ does not have 5 as one of the prime factors. Hence, $$6^n$$ can never end with a digit 0 for any natural number n.

### Question - 6

#### Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers. Solution:

Composite Numbers : Numbers which have factors other then 1 and itself.

7 × 11 × 13 + 13
= 13 × (7 × 11 + 1) = 13 × 78 is a composite number because it has factors other than 1 and itself.

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
= 5 × (7 × 6 × 4 × 3 × 2 × 1 + 1)
= 5 × (1008 + 1)
= 5 × 1009 is a composite number because it has factors other than 1 and itself.

### Question - 7

#### 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? Solution:

We will find the LCM to find the time for them to meet again at the starting point.

By prime factorization,
12 = $$2\times 2\times 3 = 2^2\times 3$$
18 = $$2\times 3\times 3 = 2\times3^2$$
LCM(12,18) = $$2^2\times 3^2 = 36$$

They will meet again at starting point after 36 minutes.