Can the number 3 to the power of n ,n being a natural number ends with 5?Give reasons
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Answer:
Step-by-step explanation:
on the factorization of 3 there was not any 5. So it cannot give a number which ends with a digit 5. Because it cannot be written in the form 2 raise to n into 5 raise to m.
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method 1
3^0=1
3^1=3
3^2=9
3^3=27
3^4=81.......
thus 3^n can't end with digit 5
method 2
3 is a prime no
prime factorisation is (3×1)
to end with no 5 then 3 shld be a multiple of 5
due to the uniqueness of fundamental therium of arithmetic s then prime factorisation is unique
thus 3^n can't end with digit 0
3^0=1
3^1=3
3^2=9
3^3=27
3^4=81.......
thus 3^n can't end with digit 5
method 2
3 is a prime no
prime factorisation is (3×1)
to end with no 5 then 3 shld be a multiple of 5
due to the uniqueness of fundamental therium of arithmetic s then prime factorisation is unique
thus 3^n can't end with digit 0
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