Show that 9^n can never end with digit zero for any natural number n
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this is because 9 square,cube,power4,5,67,8,9,10 .........................can't result into zero in last
Dhruv00:
you have to prove that
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If 9^n ended with the digit 0 then it would be divisible by 5 . This means that the prime factorization of 9^n would contain the prime factor 5.
But 9^n = 3(3^n) . This means that 9^n contains only the prime factor three which is guaranteed by the uniqueness of the fundamental theorem of arithmetic .
Hence since 9^n does not have 5 as a factor it cannot end with the digit 0.
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But 9^n = 3(3^n) . This means that 9^n contains only the prime factor three which is guaranteed by the uniqueness of the fundamental theorem of arithmetic .
Hence since 9^n does not have 5 as a factor it cannot end with the digit 0.
Please mark as brainliest if it was helpful :)
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