Exercise 3.5

			60
	6	×			10
2	×	?		5	×	?

Write the missing numbers : 6 = 2 ×

Write the missing numbers : 10 = 5 ×

		60
?	×			30
		?	×		10
				?	×	?

Write the missing numbers : 60 = × 30

Write the missing numbers : 30 = × 10

Write the missing numbers : 10 = ×

2. Which factors are not included in the prime factorisation of a composite number?

Determine the Prime factorization of numbers:

prime - and itself.

Composite - It has more than factors.

Hence, and the number itself factors are not included in the prime factorization of a composite number.

Write the smallest 5-digit number and express it in terms of its prime factors.

Instructions

Write the smallest 5-digit number

The smallest 5-digit number is .
10000 is . Thus, it will be divisible by .
The prime factorisation will be continued until the number is no longer divisible by 2.
625 divisible by 2 so, let's check with a different number: .
Now, continue until we reach 1.
Thus, the prime factorization of 10000 is as given.

5. Find all the prime factors of 1729 and arrange them in ascending order. Now state the relation, if any; between two consecutive prime factors.

Instructions

check with 1 to 10 numbers which can be divisible.

1729 / = 247.

247 is not divisible by 7 then check with 10 to 20.

247 / = 19.

19 is a prime number.

Thus, the prime factorization of 1729 is: x x .

Arranging the prime factors in ascending order: , , .

To find the relationship between two consecutive prime factors: 13 - 7 = , 19 - 13 = .

Thus, the difference between each pair of consecutive prime factors is 6. This indicates that the prime factors of 1729 are in an arithmetic progression with a common difference of 6.

6. The product of three consecutive numbers is always divisible by 6. Verify this statement with the help of some examples.

Instructions

To verify that the product of three consecutive numbers is always divisible by 6, we need to understand that 6 is the product of the prime numbers and .

Thus, for a product to be divisible by 6, it must be divisible by both 2 and 3.

Example 1 : Product of 120 = × ×

E×ample 2 : Product of 504 = × ×

Both, this e×amples are divisible by 2 and 3 then it will also divisible by 6.

7. The sum of two consecutive odd numbers is divisible by 4. Verify this statement with the help of some examples.

Instructions

To verify that the sum of two consecutive odd numbers is divisible by 4, let's first consider the properties of odd numbers.

Example 1 : sum of + = 4

Example 2 : product of + = 12

Both, this e×amples are divisible by 4.

8. In which of the following expressions, prime factorisation has been done?

Instructions

24=2×3×4

56=7×2×2×2

70=2×5×7

54=2×3×9

Prime factorization

Not Prime factorization

9. 18 is divisible by both 2 and 3. It is also divisible by 2 × 3 = 6. Similarly, a number is divisible by both 4 and 6. Can we say that the number must also be divisible by 4 × 6 = 24? If not, give an example to justify your answer.

Instructions

Let us consider the numbers which are divisible by 4 : ,,16,,,28,32,,

Now let us consider the numbers which are divisible by 6 : 6, , , 24, 30, , ,48, 54, 60.

So the numbers divisible by both 4 and 6 are , .

Example: 124 = , 126 = but not 1224 =

Thus, 12 is a number that is divisible by both 4 and 6 but not by 4×6=24.

Therefore, a number being divisible by both 4 and 6 does not imply that it must be divisible by 24.

10. I am the smallest number, having four different prime factors. Can you find me.

Instructions

To find the smallest number having four different prime factors, we need to multiply the four smallest prime numbers together. The four smallest prime numbers are , , , and .

Let calculate the product ×××.

Calculating this step-by-step: × = 6

× = 30

× =

So, the smallest number having four different prime factors is:210

Playing With Numbers

Glossary

Exercise 3.5

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Playing With Numbers

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Glossary

Exercise 3.5