Question

Show that 2powern cannot end with the digit 0 or 5 for any natural number n.​

Answer 1

Answer:

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Answer 2

$2 {}^{n}$

If it ends with 0 or 5 then it is divisible by 5. That is, the prime factorisation of $2 {}^{n}$would contain the prime 5 but prime factorisation of $2 {}^{n}$is 2×1. There is no 5 in its prime factorisation. So, the uniqueness of the fundamental theorem of arithmetic guarantees that there are no other prime in the prime factorisation of $2 {}^{n}$So there is no natural no. n for which $2 {}^{n}$ends with the digit zero or five

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