Question

Assertion : the number 6n , n begin a natural number , ends with the digit 5 Reason : the number 9n can't be end with digit 0 for any natural number n
​

Answer 1

Answer:

If a number ends with 5, then it is not divisible by 2. However, 6 is divisible by 2, and so every power of 6. That's why no such power can end with 5.

Answer 2

Reason is true but assertion is false.

If we want to prove with the help of contradiction,

Let the number $6^{n}$ can end with the digit 5.

Now, to let that happen,

$6^{n}$ must be divisible by 5.

Now, we have,

$6^{n} =(2.3)^{n}$

This means that the factors that are prime are 2 and 3.

This is a contradiction and hence,  $6^{n}$ cannot end with the digit 5.

Hence, assertion is false.

For any natural number n, $9^{n}$ can never end in zero because in order for a number to finish in zero, it must be divisible by 10, which means that its factors must be 5 and 2. However, since 9 only has 3 as a factor, raising its strength won't make 5 into a factor. Thus, it can never get to a zero.

Hence, reason is true but assertion is false.

Learn more here

https://brainly.in/question/50276577

#SPJ1