Question

2. Find the HCF of the following numbers
by prime factorisation.
a) 21, 45
b) 153, 27
c) 38, 361
d) 36, 72, 84
e) 75, 25, 120 f) 112, 140, 168
9​

Answer 1

Step-by-step explanation:

a) The prime factorization of 21 is: 3 x 7

• The prime factorization of 45 is: 3 x 3 x 5
• The prime factors and multiplicities 21 and 45 have in common are: 3
• 3 is the gcf of 21 and 45
• gcf(21,45) = 3

b) Highest common factor (HCF) of 153, 27 is 9.

• HCF(153, 27) = 9

c) gcf, hcf, gcd (38; 361) = 19

greatest (highest) common factor (divisor), calculated. The numbers have common prime factors.

d) 36=2x2x3x3

• 72=2x2x2x3x3

• 84-2×2×3×7

HCF of 36, 72 and 84 is 12.

e) The prime factorization of 25 is 5².

The prime factorization of 75 is 3 x 52.

The prime factorization of 120 is 23 x 3 x 5.

We can see the hcf is 5

f) 112=2x2

• 112=2×2×2× 2 × 7

• 140 = 2 × 2 × 5 x 7

• 168 = 2 x 2 x 2 × 3 × 7

• 2 x 2 x 7 == 28

ans is 28

$\huge{ son \: prodigal}$

Answer 2

a) 21=3 x 7=21

    45=3 x 5 x 3=45

∴ HCF = 3

b)153 = 3 x 3 x 17

  27 = 3 x 3 x 3

∴ HCF = 9

C)38 = 2 x 19

   361 = 19 x 19

∴HCF = 19

d)36 = 2 x 2 x 3 x 3

  72 = 2 x 2 x 2 x 3 x 3

  84 =2 x 2 x 3 x 7

∴HCF = 12

e)75 = 3 x 5 x 5

  25 = 5 x 5

  120 = 2 x 2 x 2 x 3 x 5

∴HCF = 5

f)112 = 2 x 2 x 2 x 2 x 7

 140 = 2 x 2 x 5 x 7

 168 = 2 x 2 x 2 x 3 x 7

∴HCF = 28

PLS MARK ME THE BRAINLIEST I TYPED IT ALL BYMYSELF

:)