Question

Check that the given arrangement ${2^m}{.5^n}(m,n \in N)$ ends with which one of the following?
(A). 5
(B). 0
(C). 25
(D). 125

Answer 1

Hint- In this question, we will approach the problem in such a way that we will check for ${2^m}$ first and check it ends with which number by generalizing and then check for ${5^n}$ and then combine their result by multiplication and then arrive at the result.

Complete step-by-steps solution -
Consider ${2^m}$ and on substituting m = 1, 2, 3, 4, ……so on as $m \in N$
We get, results as ${2^1},{2^2},{2^3},{2^4},.....$ or $2,4,8,16,.....$
Here, we can say that it always ends up to be an even number.
Now, Consider ${5^n}$ and on substituting n = 1, 2, 3, 4, ……so on as $n \in N$
We get, results as ${5^1},{5^2},{5^3},{5^4},.....$ or $5,25,125,625,.....$
Here, we can say that it always ends up having 5 at unit place.
Now, when we multiply both of them, we are having 5 multiplied by an even number and we will always get 0 at the unit's place.
Hence, we can say that by this generalization that ${2^m}{.5^n}(m,n \in N)$ always ends with 0.
$\therefore $ Option B. 0 is the correct answer.

Note- For such types of questions just keep in mind that when we are multiplying an even number with 5 it will always result in 0 as we can take an example for that like when 8 is multiplied by 5, we get 0 at unit place. Basically, from this we get to know that ${2^m}{.5^n}(m,n \in N)$ ends with which digit and we can test its divisibility by using the divisibility for ${2^m}$ and ${5^n}$.