Grade - 9

(i) 135 and 225

(ii) 196 and 38220

(iii) 867 and 255

Solution:

**(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

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 \(6q + 5\).

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

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 \(m\).

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 \(9m + 8\).

(i) 140

(ii) 156

(iii) 3825

(iv) 5005

(v) 7429

Solution:

**(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 = \(17\times 19\times 23\)

(i) 26 and 91

(ii) 510 and 92

(iii) 336 and 54

Solution:

**(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.

(i) 12, 15 and 21

(ii) 17, 23 and 29

(iii) 8, 9 and 25

Solution:

**(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

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

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.

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.

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.

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