can the number 6 and n being a natural number and with the six five digit give reason
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if 6^n ends with 0, then it must have 5 as a factor. But we know that only prime factors of 6^n are 2&3.
From the fundamental theorem of arithmetic, we know that the prime factorization of every composite number is unique.
So, 6^n can never end with 0
khem51:
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