AMC 10A/12A 2024 (Problem 1/1)What is the value of 9901⋅101−99⋅101019901\cdot101-99\cdot101019901⋅101−99⋅10101?(A) 2\text{(A)}\;2(A)2(B) 20\text{(B)}\;20(B)20(C) 200\text{(C)}\;200(C)200(D) 202\text{(D)}\;202(D)202(E) 2020\text{(E)}\;2020(E)2020Check AnswerYour answer:ABCDECheckHints (2)Hint 1(9900+1)(100+1)−(99)(10000+100+1)(9900+1)(100+1)-(99)(10000+100+1)(9900+1)(100+1)−(99)(10000+100+1)Hint 2990000+9900+100+1−990000−9900−99990000+9900+100+1-990000-9900-99990000+9900+100+1−990000−9900−99=100+1−99=100+1-99=100+1−99=100−99+1=100-99+1=100−99+1=1+1=2=1+1=2=1+1=2Final Answer(A) 2