Show that 12n cannot end with digit 0r 5 for any natural number n
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Answered by
486
If 12n ends with 0 then it must have 5 as a factor.
But, 12n=(2×2×3)n which shows that only 2 and 3 are the prime factors of 12n.
Also, we know from the fundamental theory of arithmetic that the prime factorization of each number is unique.
So, 5 is not a factor of 12n.
Hence, 12n can never end with the digit 0.
But, 12n=(2×2×3)n which shows that only 2 and 3 are the prime factors of 12n.
Also, we know from the fundamental theory of arithmetic that the prime factorization of each number is unique.
So, 5 is not a factor of 12n.
Hence, 12n can never end with the digit 0.
Answered by
157
hii
in the prime factorisation of 12n:- 2n*2n*3n 5 does not occur anywhere.and for any natural no.n there should be 0 or 5 at the end inthe factorisation of 12n which is not possible. thus 12n cannot end with digit 0 or 5 for any natural no. n
hence,showed.
hope this helps you!!
in the prime factorisation of 12n:- 2n*2n*3n 5 does not occur anywhere.and for any natural no.n there should be 0 or 5 at the end inthe factorisation of 12n which is not possible. thus 12n cannot end with digit 0 or 5 for any natural no. n
hence,showed.
hope this helps you!!
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