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Number Theory Level pending

Find the sum of digits of the remainder when 4000 ! 4000! is divided by ( 1999 ! ) 2 (1999!)^{2} .


The answer is 0.

Lee Young Kyu by Lee Young Kyu , 25, Hong Kong

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1 solution

Lee Young Kyu
Aug 6, 2014

4000 C 1999 4000C1999

= 4000 ! ( 1999 ! ) ( 2001 ! ) \frac{4000!}{(1999!)(2001!)}

= 4000 ! ( 1999 ! ) 2 ( 2000 × 2001 ) \frac{4000!}{(1999!)^{2}(2000×2001)}

4000 ! ( 1999 ! ) 2 = 2000 × 2001 × 4000 C 1999 \frac{4000!}{(1999!)^{2}}=2000×2001×4000C1999

Thus 4000 ! ( 1999 ! ) 2 \frac{4000!}{(1999!)^{2}} is an integer, so answer is 0.

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Same way.... :)

Krishna Ar - 6 years, 9 months ago

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