Question

check whether 26^n ends with the digit 5 for any natural number​

Answer 1

First let me solve one by one.

So taking the first case of 26^{n} ending with 5

So we know that for a number to end with 5, the prime factorisation of a number must have 5 as it's prime factor.

So calculating prime factorisation of 26 we get,

26 = 13 × 2

Since 26 has no prime factor 5 in it's prime factorisation, there is no natural number ' n ' for which 26^{n} will end with zero. This is guaranteed by the uniqueness of Fundamental Theroem Of Arithmetic.

Second case : 2^{n} ending with 6

For a number to end with 6, the prime factorisation of that number must be having both 2. But the prime factorisation of 2 is 2. Since it has 2 as it's factor, there is a natural number " n " for which 2^{n} will end with zero. This can be said by the uniqueness of the Fundamental Theorem of Arithmetic

Third case : Prediction of the unit's digit for the number with expansion

( 191 ) ¹⁰⁰⁹

We know that the number 1 raised to any power gives 1 as the answer in the unit's place. That is,

1^{n} will always have 1 at it's unit place.

Here ' n ' is any natural number

Hence we can say that,

( 191 ) ¹⁰⁰⁹ will have 1 at it's unit's place