911+119=(10−1)11+(10+1)9 =11C01011+11C1109+...11C11 +9C01010+9C1108+...+9C9 Sum of last two terms of both expansion as other term have at least 102 as a factor (110−1)+90+1=200 ∴911+119= Multiple of 100+200= Multiple of 100 Since 200=2×102 So, 911+119 is divisible by 102 ∴k=2
