A) True
B) False
C) Can't say
D) Partially True/False
Correct Answer: B
Solution :
| Given, n is a natural number. Let us assume dial \[{{4}^{n}}\]ends with 0, thus \[{{4}^{n}}\] is divisible by 2 and 5 both. |
| But prime factors of 4 are \[2\times 2\]. |
| \[\therefore {{4}^{n}}={{\left( 2\times 2 \right)}^{n}}={{2}^{2n}}\] |
| Thus. prime factorisation of \[{{4}^{n}}\] does not contain 5. So, the fundamental theorem of arithmetic guarantees that there are no other primes in the factorisation of \[{{4}^{n}}\]. |
| So, our assumption that \[{{4}^{n}}\]ends with 0 is wrong. |
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