IPMAT Indore 2025Number System > EasyEntered answer:✅ Correct Answer: 11Related questions:The remainder when 1!+2!+3!+...+95!1! + 2! + 3! + ... + 95!1!+2!+3!+...+95! is divided by 151515 isA polynomial P(x)P(x)P(x) leaves a remainder 222 when divided by (x−1)(x - 1)(x−1) and a remainder 111 when divided by (x−2)(x - 2)(x−2) The remainder when P(x)P(x)P(x) is divided by (x−1)(x−2)(x - 1)(x - 2)(x−1)(x−2) isThe polynomial 4x10−x9+3x11−5x7+cx6+2x5−x4+x3−4x2+6x−24x ^ {10} - x ^ 9 + 3x ^ {11} - 5x ^ 7 + c x ^ 6 + 2x ^ 5 - x ^ 4 + x ^ 3 - 4x ^ 2 + 6x - 24x10−x9+3x11−5x7+cx6+2x5−x4+x3−4x2+6x−2 when divided by x−1x - 1x−1 leaves a remainder 2.2.2. Then the value of c+6c + 6c+6 is