Exercise 1.2
6. Express each of the following numbers as a product of its prime factors. (i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429
(i)
(i) 140
Solution
Prime factors of 140 =
(ii)
(ii) 156
Solution
Prime factors of 3825 =
(iii)
(iii) 3825
Solution
Prime factors of 7429 =
(iv)
(iv) 5005
Solution
Prime factors of 5005 =
(v)
(v) 7429
Solution
Prime factors of 7429 =
2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.
(i)
(i) 12, 15 and 21
Solution
LCM × HCF = Product of the numbers.
[Note: give the numbers in ascending order only]
Prime factors of 12 =
Prime factors of 15 =
Prime factors of 21 =
HCF of 12, 15 and 21 =
LCM of 12, 15 and 21 = 2 × 2 × 3 × 5 × 7 =
Product of these three numbers = 12 × 15 × 21 =
LCM × HCF = 420 × 3 =
Note: For three numbers, LCM × HCF may not always equal the product of the numbers.
(ii)
(ii) 17, 23, and 29
Solution
[Note: give the numbers in ascending order only]
Prime factors of 17 =
Prime factors of 23 =
Prime factors of 29 =
HCF of 17, 23 and 29 =
LCM of 17, 23 and 29 = 17 × 23 × 29 =
Product of these three numbers = 17 × 23 × 29 =
LCM × HCF = 11339 × 1 =
Thus, the product of the numbers = LCM × HCF = 11339.
(iv)
(iv) 72 and 108
Solution
[Note: give the numbers in ascending order only]
Prime factors of 72 =
Prime factors of 108 =
HCF of 72 and 108 = 2 × 2 × 3 × 3 =
LCM of 72 and 108 = 2 × 2 × 2 × 3 × 3 × 3 =
Product of these two numbers = 72 × 108 =
LCM × HCF = 216 × 36 =
Thus, the product of the numbers = LCM × HCF = 7776.
(v)
(v) 306 and 657
Solution
[Note: give the numbers in ascending order only]
Prime factors of 306 =
Prime factors of 657 =
HCF of 306 and 657 = 3 × 3 =
LCM of 306 and 657 = 2 × 3 × 3 × 17 × 73 =
Product of these two numbers = 306 × 657 =
LCM × HCF = 6696 × 9 =
Thus, the product of the numbers = LCM × HCF = 60264.
3. Check whether
For a number to end with the digit 0, it must be divisible by
That means it must have both 2 and
Prime factorization of
There is no factor of
Therefore,
4. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
1014 is divisible by
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 =
Therefore, both numbers are composite because they have divisors other than
5. How will you show that
Let’s factor the expression:
(17 × 11 × 2) + (17 × 11 × 1 × 5) = 17 × 11 × 2 + 17 × 11 × 5
Take out the common factor 17 × 11:
= 17 × 11 × (2 + 5) = 17 × 11 ×
Since
it is a composite number.
6. Which digit would occupy the units place of
Let’s look for a pattern in the units digit of powers of 6:
So, the units digit of
Therefore, the units digit of