Why the number 4^n , where n is a natural number , cannot end with 0 ?
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Answered by
87
We can write 4^n as 2^2n.
Hence it has only on prime factor that is 2.
But for any number to end with zero it has to be the multiple of 5 and 2 both.
Hence it can't end with zero.
Hope it helps....pls mark brainliest.....
Hence it has only on prime factor that is 2.
But for any number to end with zero it has to be the multiple of 5 and 2 both.
Hence it can't end with zero.
Hope it helps....pls mark brainliest.....
Answered by
73
given" n" is a rational number.
Let 4^n ends with zero. then 4 is divisible by 5 but prime factors of 4 are 2 × 2.
4^n= (2×2)^n=2^2n
thus prime factorization of 4^n does not contain 5 .so the uniqueness of the fundamental theorem of arithmetic guarantees that there are no other primes in the factorization of 4^n.
hence there is no natural number"n" for which 4^n ends with the digit zero.
Let 4^n ends with zero. then 4 is divisible by 5 but prime factors of 4 are 2 × 2.
4^n= (2×2)^n=2^2n
thus prime factorization of 4^n does not contain 5 .so the uniqueness of the fundamental theorem of arithmetic guarantees that there are no other primes in the factorization of 4^n.
hence there is no natural number"n" for which 4^n ends with the digit zero.
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