Question

check whether (28)n can rnd with zero for any natural number

Answer 1

Answer:

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Step by step solution:

$we \ have \ to \ check \ 28^{n} \ end \ with \ digit \ zero \ for \ any \ natural \ number.\\\\ To \ ends \ with \ digit \ 0, \ it \ should \ be \ divisible \ by \ (2 \times5) \ for \ any \ natural \ no.\\\\but \ here \ 28^{n}=(4 \times7)^{n}\\\\so \ it \ will \ not \ end \ with \ for \ any \ natural \ number.$

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