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| 1. |
For any posutve integer n Euclid division lemma to prove that n cube-n by 6 |
| Answer» Let :-a = n³ - n= n(n² - 1)\xa0= n(n - 1)(n + 1)= (n - 1)n (n + 1)1) Now out of three (n - 1),n and (n + 1) one must be even so a is divisible by 22) Also (n - 1),n and (n + 1) are three consecutive integers thus as proved a must be divisible by 3\xa0From (1) and (2)a must be divisible by 2 × 3 = 6Hence n³ - n is divisible by 6 for any positive integer n | |