### Chapter 1: Real Numbers

Q
##### Real Numbers Solutions

Question:

Check whether $6^{n}$ can end with the digit 0 for any natural number n.

To find whether the number ends with a digit 0 for any natural number n we first find the factorization of the value.

We know that the factors for 6 = 2 x 3

So,     6= 2n x 3n

Thereby putting values like positive natural numbers in n = 1, 2, 3….. n but to get a 0 at the end the number has to be either multiplied by 0 or by 5.

Otherwise, the value of 2n  and 3n will never end with 0. The unit value of 2n will always be even number except for zero and unit value of 3n will be 3, 9, 7, 1 thereby multiplying both factors will never have zero at the unit place of a number.

Therefore, there is no value of n that will give unit digit of equal to zero.

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