( 1 0 0 ! − 9 9 ! ) × ( 9 8 ! − 9 7 ! ) × … ( 2 ! − 1 ! ) ( 1 0 0 ! + 9 9 ! ) × ( 9 8 ! + 9 7 ! ) × … ( 2 ! + 1 ! )
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Can you elaborate on the second to last line?
I have to say this- This is a very elegant problem, so much fun to solve when so many terms cancel out!
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Yes it is, BTW this is not my original idea, it is just a variation from the inspiration because someone could have brute-forced that problem
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I actually first solved the inspiration with brute force only, but then realized how weird my thinking was and attempted yours by myself!
Awesome problem! Awesome solution! Awesome Everything! I love it when something in math looks really complex until you simplify it. This is the beauty of math. Awesome!
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Thank you Yashvardhan. Maths is very beautiful indeed
By the way, what does using brute-force to solve a problem mean?? You guys use it a lot.
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Checking it out every option / case there could be
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= ( 1 0 0 ! − 9 9 ! ) × ( 9 8 ! − 9 7 ! ) × … ( 2 ! − 1 ! ) ( 1 0 0 ! + 9 9 ! ) × ( 9 8 ! + 9 7 ! ) × … ( 2 ! + 1 ! ) = ( 1 0 0 × 9 9 ! − 9 9 ! ) × ( 9 8 × 9 7 ! − 9 7 ! ) × … ( 2 × 1 ! − 1 ! ) ( 1 0 0 × 9 9 ! + 9 9 ! ) × ( 9 8 × 9 7 ! + 9 7 ! ) × … ( 2 × 1 ! + 1 ! ) = 9 9 ! ( 1 0 0 − 1 ) 9 9 ! ( 1 0 0 + 1 ) ⋅ 9 7 ! ( 9 8 − 1 ) 9 7 ! ( 9 8 + 1 ) ⋯ 9 7 ! ( 2 − 1 ) 1 ! ( 2 + 1 ) = 9 9 1 0 1 ⋅ 9 7 9 9 ⋯ 1 3 = 1 0 1