Find the last \(3\) digits of \(101^{101}\).
1012 = 10201
1013 = 1030301
1014 = 104060401
continuing the pattern 101101 last 3 digits would be 101
To confirm EP's answer....look at the last two terms of the binomial expansion of (101)^101 = (100 + 1)^101 =
C(101, 100) * 100 * 1^100 + C(101, 101) * 100^0 * 1^101 =
101 * 100 + 1 * 1 * 1 =
10100 + 1 =
10101