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Consider the number of $ {4^n} $ , where $ n $ is a natural number. Check whether there is any value of $ n \in N $ for which $ {4^n} $ ends with the digit zero.
A.Yes
B.No
C.Ambiguous
D.Data insufficient

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Last updated date: 13th Jul 2024
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Answer
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Hint: In order to check whether the given value ends with the digit zero or not, factorize the value given and check for the factors, if the value contains $ 2 $ and $ 5 $ as their factors then it will give its end digit as zero, but if only one among these factors present then it will not give the last digit as zero.

Complete step-by-step answer:
We are given with the number $ {4^n} $ , where $ n $ is the exponential power of $ 4 $ .
Since, we know that a number to end with zero or a number whose last digit is zero must be divisible by $ 2 $ and $ 5 $ .
For example, let’s take an example of some numbers that have numbers whose last digit is $ 0 $ .
 $ 10 = 2 \times 5 $
 $ 100 = 2 \times 2 \times 5 \times 5 $
Hence, the numbers that end with the digit $ 0 $ contain the factors $ 2 $ and $ 5 $ .
But we are given with the value $ 4 $ , writing its factor and we get:
 $ 4 = 2 \times 2 $
And, for $ {4^n} $ it can be expanded as $ {4^n} = {\left( {2 \times 2} \right)^n} $ .
Since, the power of $ 2 $ are same, so it can be expanded as $ {4^n} = {\left( {2 \times 2} \right)^n} = {2^n} \times {2^n} $ .
But we are only getting $ 2 $ as a factor and $ 5 $ is not present in the expansion of factors, that means $ {4^n} $ cannot end with $ 0 $ . As, to get the last digit zero both the factors $ 2 $ and $ 5 $ should be present.
Therefore, the Correct Option is B that is No (there is no value of $ n \in N $ for which $ {4^n} $ ends with the digit zero).
So, the correct answer is “Option B”.

Note: Prime numbers are numbers which can be divided only by the number $ 1 $ and itself, or we can say that the number should have only $ 1 $ and itself as its factors.
The number whose last digit is zero, must be divisible by $ 2 $ or $ 5 $ .