Let's say MN intersect AX at O , MO = 1/2 BX ∆AMO and ∆ABX are similar triangles so Area∆ AMO/ Area∆ABX = MO²/BX² = 1/4 So if Area of ∆AMO =...

Hello Abhi , 1021^1022 mod 1023 Using Negative remainder concept (-2)^1022 mod 1023 2^1023 mod 1023 2^10 mod 1023 = 1024 mod 1023 = 1 ...

Hello , Ritika 199 ! = 1× 2 × 3 × ......67 × 68 × 69 ×........132 × 133×134×......199. 1× 2 × 3 × ......67 × 68 × 69 ×........132 × 133×134×.....

Hello Kinshuk ! a^19 + 17a - 19 = 0 a^19 = 19 -17a similarly , b^19 = 19 - 17b , c^19 = 19 - 17c ..... So a^19 + b^19 + c^19 +....... = 19 - 17a + 1...

Alternate Approach