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What is remainder from division of $A$ by $10$? $A=0!+3!+6!+9!+…+81!$

I thought I was kind of expert in high school mathematics until I was hit in punch by my brother who is in secondary school. Guys I will really appreciate if anyone can get me out of this mess.

1ENİGMA1
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Reshad Safi
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1 Answers1

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$6!, 9!, 12!, ... 81!$ will all end in 0 since they each a 5 and a 2 as factors. So to find the units place, just look at $0! = 1$ and $3! = 6$. So your units place for the sum will be $1 + 6 + 0 + 0 + 0 + ... + 0 = 7$. And there's your remainder.

Kurt Schwanda
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  • I can't believe how easy it was. I spent almost 2 hours to find sum of the sequence while I didn't even need it. Thank you very much – Reshad Safi Sep 25 '18 at 14:06
  • Is there any specific formula for Sum of such factorial sequences? – Reshad Safi Sep 25 '18 at 14:08
  • You can check this out: https://math.stackexchange.com/questions/227551/sum-k-1-2-3-cdots-n-is-there-a-generic-formula-for-this But I don't think you'll find a formula that's nearly as simple as it is for an arithmetic series or a sum of squares! The terms get very big, very quickly. Google's telling me 81! = 5.797126e+120. I don't think you'll be asked for the sums of many factorials. This was a bit of a trick question trying to get you to notice that for every n >= 5, n! will be a multiple of 10. – Kurt Schwanda Sep 25 '18 at 14:18