
The number of pairs of positive integers \[\left( {m,{\text{ }}n} \right)\] such that ${m^n} = 25$ is:
(A) 0
(B) 1
(C) 2
(D) more than 2
Answer
559.8k+ views
Hint: We know that 25 is the square of 5 i.e. ${5^2} = 25$. And 5 is a prime number so it cannot be written as a multiple of two positive integers except 1 and 5 $\left( {5 = 1 \times 5} \right)$. Use this result to find the required pair of positive integers satisfying the given condition.
Complete step-by-step answer:
According to the question, we have to find the number of pairs of positive integers \[\left( {m,{\text{ }}n} \right)\] satisfying the condition ${m^n} = 25$.
We know that 25 is a whole square i.e. square of a positive integer 5. This can be shown as:
$ \Rightarrow {5^2} = 25$
Comparing this with ${m^n} = 25$, we have $m = 5$ and $n = 2$. Therefore \[\left( {5,{\text{ }}2} \right)\] is a pair satisfying the given condition.
Now, it is evident that both $m$ and $n$ cannot be greater than 25 but $m$ can be 25 in one case as shown:
$ \Rightarrow {25^1} = 25$
Again in comparison, we have $m = 25$ and $n = 1$. Thus \[\left( {25,{\text{ }}1} \right)\] is also a valid pair.
Further we know that 25 is not a whole cube and it cannot be written as a positive integer to power another positive integer except ${5^2} = 25$ and ${25^1} = 25$.
Hence \[\left( {5,{\text{ }}2} \right)\] and \[\left( {25,{\text{ }}1} \right)\] are the only valid pairs satisfying the given condition. (C) is the correct option.
Note:
25 can also be written as the square of -5 as shown below:
$ \Rightarrow {\left( { - 5} \right)^2} = 25$
But we have to consider only positive integers for $m$ and $n$ that’s why this case cannot give us a valid pair and hence it can be neglected.
Complete step-by-step answer:
According to the question, we have to find the number of pairs of positive integers \[\left( {m,{\text{ }}n} \right)\] satisfying the condition ${m^n} = 25$.
We know that 25 is a whole square i.e. square of a positive integer 5. This can be shown as:
$ \Rightarrow {5^2} = 25$
Comparing this with ${m^n} = 25$, we have $m = 5$ and $n = 2$. Therefore \[\left( {5,{\text{ }}2} \right)\] is a pair satisfying the given condition.
Now, it is evident that both $m$ and $n$ cannot be greater than 25 but $m$ can be 25 in one case as shown:
$ \Rightarrow {25^1} = 25$
Again in comparison, we have $m = 25$ and $n = 1$. Thus \[\left( {25,{\text{ }}1} \right)\] is also a valid pair.
Further we know that 25 is not a whole cube and it cannot be written as a positive integer to power another positive integer except ${5^2} = 25$ and ${25^1} = 25$.
Hence \[\left( {5,{\text{ }}2} \right)\] and \[\left( {25,{\text{ }}1} \right)\] are the only valid pairs satisfying the given condition. (C) is the correct option.
Note:
25 can also be written as the square of -5 as shown below:
$ \Rightarrow {\left( { - 5} \right)^2} = 25$
But we have to consider only positive integers for $m$ and $n$ that’s why this case cannot give us a valid pair and hence it can be neglected.
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