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Question 11
Show that 12n cannot end with the digit 0 or 5 for any natural number n.


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Solution

If any number ends with the digit 0 or 5, it is always divisible by 5.
If 12n ends with the digit zero or five, it must be divisible by 5.
This is possible only if prime factorization of 12 contains the prime number 5.
Now, 12=2×2×3=22×3
12n=(2×2×3)n=2n×2n×3n
12n is not divisible by 5 as 5 is not in the prime factorisation of 12n.
Hence, there is no value of n ϵ N for which 12n ends with digit zero or five.


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