Show that 5n cannot end with the digit 2 for any natural number n
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IF THE NUMBER 5n, FOR ANY n, WERE TO BE DIVISIBLE BY 2,THEN IT WOULD HAVE 2 IN ITS PRIME FACTORISATION.
THAT IS, THE PRIME FACTORISATION OF 5nWOULD CONTAIN THE PRIME NO 2.THIS IS, NOT POSSIBLE BECAUSE FACTORS OF 5n IS (5)n
SO, THE PRIME FACTORISATION OF 5nCONTAINS ONLY AND ONLY 5 NOT 2. SO THE UNIQUENESS OF THE FUNDAMENTAL THEOREM OF ARITHMETIC GURANTEES THAT THERE ARE NO OTHER PRIMES IN THE FACTORISATION OF 5n.
SO, THERE IS NO NATURAL NUMBER n FOR WHICH 5n IS DIVISIBLE BY 2
THAT IS, THE PRIME FACTORISATION OF 5nWOULD CONTAIN THE PRIME NO 2.THIS IS, NOT POSSIBLE BECAUSE FACTORS OF 5n IS (5)n
SO, THE PRIME FACTORISATION OF 5nCONTAINS ONLY AND ONLY 5 NOT 2. SO THE UNIQUENESS OF THE FUNDAMENTAL THEOREM OF ARITHMETIC GURANTEES THAT THERE ARE NO OTHER PRIMES IN THE FACTORISATION OF 5n.
SO, THERE IS NO NATURAL NUMBER n FOR WHICH 5n IS DIVISIBLE BY 2
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if the number 5 and for any and word to be divisible by 2 then it would have to in its prime factorization that is the prime factorization of 5 and what contain the prime number 2 this is not possible because factors of 5 and is 5 and soap prime factorization of 5 and contains only and only fine not to
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