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ParthKohli

  • 3 years ago

How will we solve this? What will be the remainder after we divide 50! + 49! + 48!.......2! + 1! by 5! ?

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  1. ParthKohli
    • 3 years ago
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    @FoolForMath I need your genius here

  2. ash2326
    • 3 years ago
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    50!.... 5! each of these have a 5! so remainder will be 4!+3!+2!+1!=24+6+2+1=33

  3. ParthKohli
    • 3 years ago
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    I am not sure how you got it :|

  4. FoolForMath
    • 3 years ago
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    It's easy, Starting from 5! every factorial is a divisible by 5.

  5. FoolForMath
    • 3 years ago
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    So, just ignore the rest.

  6. ParthKohli
    • 3 years ago
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    Oh yes!

  7. ash2326
    • 3 years ago
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    Did you understand @ParthKohli ?

  8. ParthKohli
    • 3 years ago
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    Yes! that's the principle of how we divide \(\Large {6! \over 5!}\) and we get \(6\)... so 6 is the quotient and there is no remainder! ^.^

  9. ademm37
    • 3 years ago
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    |dw:1340545398596:dw|

  10. mukushla
    • 3 years ago
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    how about equation 1!+2!+3!+...+n!=n^2 how many answers does it have n is nature?

  11. ParthKohli
    • 3 years ago
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    Thank you all!

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