Question

Show that 7" can not end with the digit o for any natural number n.​

Answer 1

Solution :

If 7ⁿ ends with a 0 the it must be divisible by 10 i.e., its prime factors should have the factors of both 2 and 5 since 2×5=10. Therefore by the fundamental theorem of arithmetic (there is no prime factors of 7 other than 7 and 1) we can conclude that 7ⁿ can not end with 0.

Answer 2

Solution :

If 7ⁿ ends with a 0 the it must be divisible by 10 i.e., its prime factors should have the factors of both 2 and 5 since 2×5=10. Therefore by the fundamental theorem of arithmetic (there is no prime factors of 7 other than 7 and 1) we can conclude that 7ⁿ can not end with 0.

Answer 3

Solution :

If 7ⁿ ends with a 0 the it must be divisible by 10 i.e., its prime factors should have the factors of both 2 and 5 since 2×5=10. Therefore by the fundamental theorem of arithmetic (there is no prime factors of 7 other than 7 and 1) we can conclude that 7ⁿ can not end with 0.

Answer 4

Solution :

If 7ⁿ ends with a 0 the it must be divisible by 10 i.e., its prime factors should have the factors of both 2 and 5 since 2×5=10. Therefore by the fundamental theorem of arithmetic (there is no prime factors of 7 other than 7 and 1) we can conclude that 7ⁿ can not end with 0.