Maths Olympiad Prep

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Algebra Difficulty 5.1 AIME, harder Find the answer

The expression ⌊x⌋\lfloor x\rfloor denotes the greatest integer less than or equal to xx. Find the value of ⌊2002!2001!+2000!+1999!+⋯+1!⌋.\left\lfloor\frac{2002!}{2001!+2000!+1999!+\cdots+1!}\right\rfloor.

A number or a short expression. Spacing and $ signs are ignored.

Solution

2000 We break up 2002! = 2002(2001)! as 2000(2001!)+2⋅2001(2000!)=2000(2001!)+2000(2000!)+2002⋅2000(1999!)>2000(2001!+2000!+1999!+⋯+1!)2000(2001!)+2 \cdot 2001(2000!)=2000(2001!)+2000(2000!)+2002 \cdot 2000(1999!) >2000(2001!+2000!+1999!+\cdots+1!) On the other hand, 2001(2001!+2000!+⋯+1!)>2001(2001!+2000!)=2001(2001!)+2001!=2002!2001(2001!+2000!+\cdots+1!)>2001(2001!+2000!)=2001(2001!)+2001!=2002! Thus we have 2000<2002!/(2001!+⋯+1!)<20012000<2002!/(2001!+\cdots+1!)<2001, so the answer is 2000.

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