Question

Show that any number of the form 6n where n is a natural number can never end with the digit 0.

Please give me answer quickly.

Answer 1

prime factorization of 6=(3×2)^n

for any number to end withzero the prime factorization should contain 2 and 5 but 6 does not contain 2 and 5 in its prime factorization

hence it can never end with digit 0

Answer 2

Correct question:

Show that any number of the form 6^n where n is a natural number can never end with the digit 0.

Explanation:

• METHOD 1 If any digit has the remaining digit 10 which means it divisible through 10.

The issue of 10=2×5,

So the value of 6^n

ought to be divisible through 2 and 5.

Both 6 ^n

is divisible through 2 however now no longer divisible through 5.

• Method 2 :

The top factorisation of

$6^n= (2×3)^n$

• So it does comprise a 2 in its top factorisation however in this situation a five is missing .
• That method that it isn't divisible via way of means of 2×five each i.e. 10, So it isn't viable for the wide variety to stop with a 0 if its now no longer divisible via way of means of 10.
• This is the easy manner to country that “ for any actual wide variety to stop with 0 its top factorisation should comprise two and five in it.”

So to our case the wide variety 6^n will by no means stop with 0 for any original value of 10.

(#SPJ2)