Question

show 32^n ends not with 0 where n natural number
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Answer 1

Answer:

So the answer will be 1

Answer 2

Answer:

Proof is given below.

Step-by-step explanation:

In this proof the Fundamental Theorem of Arithmetic is used, which states that there is only one way to uniquely write the prime factors of a number, except the order in which the primes are written.

Eg: 30 = 2 × 5 × 3 (This is the only way to prime factorise 30)

     30 = 5 × 3 × 2 (Same primes but written in different order)

     30 = 3 × 2 × 5 (Same primes but written in different order)

Now, $32^n$ = $(2^5)^n$ = ${2}^{5n}$ = 2 × 2 × 2 × 2 ... (5n times)

But, any number that ends with zero must have a prime factor of 5.

Eg: 30 = 2 × 3 × 5

     40 = $2^3$ × 5

However, the prime factorisation of $32^n$ contains only 2s and no 5s. The uniqueness of the Fundamental Theorem of Arithmetic guarantees that there is no other way to write the prime factorisation. So $32^n$ cannot end with zero for n ∈ N.

Hence proved.