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May 8th, 2014, 05:54 AM   #1
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Remainder of (1!+2!+....99!+100!)/24

Find the remainder when 1!+2!+3!+4!+....................+96!+97!+98!+99!+1 00! is divided by 24.

I know 24=4!.
Hence we can write

remainnder when 1!+2!+3!+4!+....................+96!+97!+98!+99!+1 00! is divided by 4!.

I don't have any idea how to proceed. But I am sure that there must be a shortcut instead of finding 100!, 99! like that and summing them up.
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May 8th, 2014, 06:09 AM   #2
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for every n>=4 n! is divisible by 24 hence the remainder is just 1!+2!+3!=9
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May 8th, 2014, 06:17 AM   #3
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Every term after 3! is a multiple of 24 so we can write

$$ \frac{1! + 2! + 3! + 24k}{24} = \frac{9 + 24k}{24} $$

The numerator is 9 more than a multiple of 24 therefore the remainder is 9 after division by 24.

Answer: 9

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May 8th, 2014, 06:18 AM   #4
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1 2 6 24 120 .....

24 120 and on all divisible by 24

First 3: 1+2+6 = 9 ; 9/24 = .375

So remainder will be [integer.375] where integer is a huge mudder: 157 digits

That's all you get from me!
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