Mathematics
Find the HCF of the given numbers by prime factorisation method:
(i) 28, 36
(ii) 54, 72, 90
(iii) 105, 140, 175
Answer
(i) 28, 36
| 2 | 28 |
|---|---|
| 2 | 14 |
| 7 | 7 |
| 1 |
So, 28 = 2 × 2 × 7.
| 2 | 36 |
|---|---|
| 2 | 18 |
| 3 | 9 |
| 3 | 3 |
| 1 |
So, 36 = 2 × 2 × 3 × 3.
The common prime factor is 2, occurring twice in both numbers.
Hence, HCF of 28 and 36 = 2 × 2 = 4.
(ii) 54, 72, 90
| 2 | 54 |
|---|---|
| 3 | 27 |
| 3 | 9 |
| 3 | 3 |
| 1 |
So, 54 = 2 × 3 × 3 × 3.
| 2 | 72 |
|---|---|
| 2 | 36 |
| 2 | 18 |
| 3 | 9 |
| 3 | 3 |
| 1 |
So, 72 = 2 × 2 × 2 × 3 × 3.
| 2 | 90 |
|---|---|
| 3 | 45 |
| 3 | 15 |
| 5 | 5 |
| 1 |
So, 90 = 2 × 3 × 3 × 5.
The common prime factors are 2 (one time) and 3 (two times) in all three numbers.
Hence, HCF of 54, 72 and 90 = 2 × 3 × 3 = 18.
(iii) 105, 140, 175
| 3 | 105 |
|---|---|
| 5 | 35 |
| 7 | 7 |
| 1 |
So, 105 = 3 × 5 × 7.
| 2 | 140 |
|---|---|
| 2 | 70 |
| 5 | 35 |
| 7 | 7 |
| 1 |
So, 140 = 2 × 2 × 5 × 7.
| 5 | 175 |
|---|---|
| 5 | 35 |
| 7 | 7 |
| 1 |
So, 175 = 5 × 5 × 7.
The common prime factors are 5 (one time) and 7 (one time) in all three numbers.
Hence, HCF of 105, 140 and 175 = 5 × 7 = 35.
Related Questions
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(i) 198, 429
(ii) 20, 64, 104
(iii) 120, 144, 204
Fill in the blanks:
(i) HCF of two consecutive natural numbers is ….
(ii) HCF of two consecutive odd numbers is ….
(iii) HCF of two consecutive even numbers is ….