Exercise 3.5

1. Here are two different factor trees for 60. Write the missing numbers.

(i)Factor tree (a):

It starts as \(60 = 6 \times 10\). Then: \(6 = 2 \times 3\), \(10 = 2 \times 5\). Factor tree of 60 (a)

So missing numbers are: 3, 2

(ii)Factor tree (b):

It starts as \(60 = 30 \times 2\). Then: \(30 = 10 \times 3\), then \(10 = 2 \times 5\). Factor tree of 60 (b)

So missing numbers: 2,5,3 & 2

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

Prime factorisation includes only prime numbers.

So, composite factors and 1 are not included.

Final answer : Composite factors and 1 are not included in prime factorisation.

3. Write the greatest 4-digit number and express it in terms of its prime factors.

Greatest 4-digit number = 9999

Prime factorisation of 9999 step by step:

So, \(9999 = 3 \times 3 \times 11 \times 101\)

\(9999 = 3^2 \times 11 \times 101\)

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

Smallest 5-digit number = 10000

Prime factorisation:

\(10000 = 2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 5 \times 5\)

\(10000 = 2^4 \times 5^4\)

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.

Prime factorisation of 1729:

So, \(1729 = 7 \times 13 \times 19\)

So, prime factors: 7, 13, 19

In ascending order: 7, 13, 19

Difference between consecutive factors:
\(13 - 7 = 6\), \(19 - 13 = 6\)

This is the famous Hardy-Ramanujan number! The prime factors are in arithmetic progression with common difference 6.

Prime factors: 7, 13, 19. Consecutive factors differ by 6.

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

Three consecutive numbers like: (1, 2, 3) or (4, 5, 6) etc.

Lets find product and check:

\( 1 \times 2 \times 3\) = 6 (Divisible by 6)

\( 4 \times 5 \times 6\) = 120 (Divisible by 6)

\( 7 \times 8 \times 9\) = 504 (Divisible by 6)

\( 10 \times 11 \times 12\) = 1320 (Divisible by 6)

Reason:

In any three consecutive numbers:
– At least one is even → divisible by 2
– At least one is divisible by 3
So product is divisible by \(2 \times 3 = 6\).

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

Prime factorisation means only prime numbers are multiplied.

(a) \(24 = 2 \times 3 \times 4\) → 4 is composite → ❌
(b) \(56 = 7 \times 2 \times 2 \times 2\) → all primes → ✅
(c) \(70 = 2 \times 5 \times 7\) → all primes → ✅
(d) \(54 = 2 \times 3 \times 9\) → 9 is composite → ❌

Prime factorisation done in: (b) and (c)

8. 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.

No, we cannot say that.

Example: 12 is divisible by both 4 and 6, but \( 4 \times 6= 24 \).

And 12 is not divisible by 24.

So, a number divisible by both 4 and 6 is not necessarily divisible by 24.

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

Take the four smallest prime numbers:
\(2, 3, 5, 7\)

Multiply them: \(2 \times 3 \times 5 \times 7 = 210\)

The smallest number is 210.