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by karthickgururaj
3155 days ago
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That leads me to think what can be even more unsatisfying proof. How about the following: It is easy to see that the last digit of x^5 is same as k^5, where k is the last digit of x. Then the problem reduces to proving that the ten possibilities, 0^5, 1^5, 2^5, ... 9^5 ends respectively in 0, 1, 2, ...9. |
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