Question

Check whether 6n can end with the digito for any natural number n.
Explain why 7*11*13+13 and 7*6*5*4*3*2*1 + 5 are composite numbers
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Answer 1

Answer:

A)

If any number ends with the digit 0, it should be divisible by 10 or in other words, it will also be divisible by 2 and 5 as 10 = 2 × 5

Prime factorisation of 6n = (2 ×3)n

It can be observed that 5 is not in the prime factorisation of 6n.

Hence, for any value of n, 6n will not be divisible by 5.

Therefore, 6n cannot end with the digit 0 for any natural number n.

B)

Given 7×11×13+13

=13×(7×11+1)=3×78  

This number is multiple of two integers.Hence it has more than two factors.Hence it is a composite number.

similarly in

7×6×5×4×3 ×2×1+5    

=5(7×6×4×3 ×2×1+1)=5×1009    

This number is multiple of two integers.Hence it has more than two factors.Hence it is a composite number.

Answer 2

Question 1:

By prime factorising 6ⁿ, we get,

⇒ 6ⁿ = (2 × 3)ⁿ

This shows that prime factors of 6 are only 2 & 3. If a number should end with digit zero, then it should be divisible by both 2 & 5. But, 5 doesn't occur in the prime factorisation of 6. So there is no natural number n for which 6ⁿ ends with the digit 0.

$\hrulefill$

Question 2:

We have,

• 7 × 11 × 13 + 13
• 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

Now,

⇒ 7 × 11 × 13 + 13

⇒ 13 × (7 × 11 + 1)

⇒ 13 × 78

Hence, according to the fundamental theorem of Arithematic, it is a composite number beacuse it can be expressed as a product of prime numbers.

Again,

⇒ 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

⇒ 5 × (7 × 6 × 4 × 3 × 2 × 1 × 1 + 1)

⇒ 5 × 1009

Hence, it is also a composite number.

Additional Information:

Fundamental theorem of Arithematic:

• Every composite number can be expressed as product of primes and this factorisation is unique apart from the order in which the prime factors occur.