Check whether 6" can end with the digit 0 for any natural number n ?
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We know that ,
2 × 5 make 0 at the end of digit but 6 can be written in 2 × 3 and 2 × 3 does not make 0 at the end of digit.
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If any digit has the last digit 10 that means it divisible by 10.
The factor of 10=2×5,
So value of 6^n should be divisible by 2 and 5.
Both 6^n is divisible by 2 but not divisible by 5.
So, it can not end with 0.
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