Question

If m and n are two natural numbers such that ${{m}^{n}}$=32, then the value of ${{n}^{mn}}$ is?

Answer 1

Hint: For solving this problem, we should know the basics of powers and roots of a number. Further, since we lack any additional type of information to solve this question algebraically or in a systematic way, we try to solve the problem by listing down the divisors of 32 and then proceeding.

Complete step by step answer:
It is given that m and n are two natural numbers such that ${{m}^{n}}$=32 and have to find a value of ${{n}^{mn}}$. Since, we have two variables m and n, but have only one equation, we have to think of a different way to solve the problem. Since, we have to represent in the form of ${{m}^{n}}$, we list down the divisors of 32. We have,

32 = $2\times 2\times 2\times 2\times 2$

Thus, we can write 32 = ${{2}^{5}}$. (m=2, n=5)

Another solution which exists is 32 = ${{32}^{1}}$ (trivial solution). Here, m = 32, n=1.

Thus, there are two cases.

Case 1: m=2, n=5

Now, we have ${{n}^{mn}}$ = ${{5}^{2\times 5}}$ = ${{5}^{10}}$= 9765625

Case 2: m=32, n=1

Now, we have ${{n}^{mn}}$ = ${{1}^{32\times 1}}$ = ${{1}^{32}}$= 1

Thus, we have two values of ${{n}^{mn}}$. These are 1 and 9765625.


Note: It is important to remember while solving the question which cannot be solved algebraically or systematically, we should be aware about including all the cases for the specific problem. Thus, it is important not to forget trivial cases such as the case 2 where m=32 and n=1 in the above problem. Further, it is generally helpful to remember the basic value of power and roots such as the values of ${{2}^{3}},{{2}^{4}},{{2}^{5}},{{3}^{3}},{{3}^{4}}$ and so on. This helps in getting a basic intuition to solve such problems much easily.