Saved Bookmarks
| 1. |
Using binomial theorem, prove that `6^n-5n`always leaves remainder 1 when divided by 25. |
|
Answer» `6^n = (1+5)^n` `= C(n,0)1^n5^0+C(n,1)1^(n-1)5^1+C(n,2)1^(n-2)5^2+C(n,3)1^(n-3)5^3+...C(n,n)1^0 5^n` `6^n=1+5n+25(C(n,2)+C(n,3)5+C(n,4)5^2+...C(n,n)5^(n-2))` `6^n-5n = 1+25k`, where ` k = C(n,2)+C(n,3)5+C(n,4)5^2+...C(n,n)5^(n-2)` As, `k` is always positive, so `25k` will always be divisible by `25`. Thus, if we divide `6^n-5n` by `25`, remainder will be `1`. |
|