#### Mathematics (NTSE/Olympiad)

# Real Numbers

__Fundamental Theorem of Arithmatic__

** Statement : Every composite number can be decomposed as a product prime numbers in a unique way, except for the order in which the prime numbers occur.**

For example :

(i) 30 = 2 × 3 × 5, 30 = 3 × 2 × 5, 30 = 2 × 5 × 3 and so on.

(ii) 432 = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 2^{4} × 3^{3} or 432 = 3^{3} × 2^{4}.

(iii) 12600 = 2 × 2 × 2 × 3 × 3 × 5 × 5 × 7 = 2^{3} × 3^{2} × 5^{2} × 7

In general, a composite number is expressed as the product of its prime factors written in ascending order of their values. Some examples are

(i) 6615 = 3 × 3 × 3 × 5 × 7 × 7 = 3^{3} × 5 × 7^{2}

(ii) 532400 = 2 × 2 × 2 × 2 × 5 × 5 × 11 × 11 × 11

**Example : Consider the number 6**^{n}, where n is a natural number. Check whether there is any value of n ∈ N for which 6^{n} is divisible by 7.

Solution: Since, 6 = 2 × 3; 6^{n} = 2^{n} × 3^{n}

↠ The prime factorisation of given number 6^{n}

↠ **6**^{n} is not divisible by 7. (Ans)

Ex.16 Consider the number 12^{n}, where n is a natural number. Check whether there is any value of n ∈ N for which 12^{n} ends with the digit zero.

Solution: We know, if any number ends with the digit zero it is always divisible by 5.

↠ If 12^{n} ends with the digit zero, it must be divisible by 5.

This is possible only if prime factorisation of 12^{n} contains the prime number 5.

Now, 12 = 2 × 2 × 3 = 2^{2} × 3

↠ 12^{n} = (2^{2} × 3)^{n} = 2^{2n} × 3^{n}

i.e., prime factorisation of 12^{n} does not contain the prime number 5.

↠ There is no value of n ∈ N for which 12^{n} ends with the digit zero. (Ans)