My Math Forum Remainder of (1!+2!+....99!+100!)/24

 Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

 May 8th, 2014, 05:54 AM #1 Newbie   Joined: May 2014 From: India Posts: 6 Thanks: 0 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.
 May 8th, 2014, 06:09 AM #2 Senior Member   Joined: Nov 2010 Posts: 288 Thanks: 1 for every n>=4 n! is divisible by 24 hence the remainder is just 1!+2!+3!=9 Thanks from agentredlum
 May 8th, 2014, 06:17 AM #3 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 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 Thanks from CRGreathouse
 May 8th, 2014, 06:18 AM #4 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 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! Thanks from GovindBalaji

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# find remainder when 1! 2! 3! .... 100! is divided by 6

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