Last Three
What are the last three 3 digits of ( 1 ! + 2 ! + 3 ! + 4 ! + ⋯ + 9 9 ! ) 1 9 9 6
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The answer is 641.
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1 solution
@Precious Prestosa -This is an excellent problem and is quite under-rated. I'm glad that my friend Jayakumar pointed out the mistake in this problem to you. Good going! BTW,You had to calculate the 1...14! sum by hand only...right? (cuz..I did so :P)
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Yes. I'm so thankful. And yes also, I did it by hand. Dunno other ways then hahaha. :)
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And my NT rating dropped by 2 points coz...I wasted two tries...Please be careful from next time XD.
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@Krishna Ar – How did you know your ratings?
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@Astro Enthusiast – I was just kidding....LOL...
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@Krishna Ar – I didn't see that coming. I thought changes are not done at the same time with all brilliant users. HAHAHA.
What was your approach @Krishna Ar ?
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Same as what precious did @Kartik Sharma - Wait, Mr.sharmaji, didnt you calculate the sum of first 14 factorials by hand....and...what....do you know to write code in python?
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@Krishna Ar – Yes, I did calculate it by hand, but why this question??? And, who told you that I know to write a code in python... yes, I do know but at the beginner level only.
Hey, scroll down and see my comment. Does that make sense??? I just wanted to know!! @Krishna Ar
You can reduce the expression even further to 3 9 6 ( 3 2 S + 1 ) modulo 1000 where S = 5 ! + 6 ! + … + 1 4 ! using binomial theorem. I leave this as an exercise.
nice problem!! well, is it a rule that 313^1996 mod 1000 is 641.
I am asking this coz I calculated it using modulo properties. I must also tell how I found it.
313^(phi(1000)) = 1 mod 1000
313^400 = 1 mod 1000
313^1600 = 1 mod 1000
Now, we have to find 313^396.
313^9 = 73 mod 1000
313^36 = 73^4 mod 1000
313^396 = 241^11 mod 1000
= 641 mod 1000
Hence, 313^(1600+396) = 1*641 mod 1000
313^1996 = 641 mod 1000
Problem is under-rated....
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Sharmaji, How did you get the crucial 313^9= 73 mod 1000 ?? Hand?
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can someone explain how to find 313^1996 last digits from the basic,i dont understand the terminology of '1 mod 1000' etc..
I have a question: how did you find out that 313^400=1 mod 1000? Coz I calculated by hand but couldn't figure out this part only.
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Euler's theorem!
The solution is incomplete. I get up to 313 to the power of 1996, just as explained here by hand, but for the mod 1000 part do we just use wolfram alpha or what? Cause that's what I did, and if the solution did not say anything about it, meaning it's OK to 'cheat a little' or even more than that?
Factorial of any number n where ( 1 5 ≤ n ) , has zeroes as its last three digits. So, ( 1 ! + 2 ! + 3 ! + 4 ! + ⋯ + 1 4 ! ) mod 1 0 0 0 is 3 1 3 . So 3 1 3 1 9 9 6 mod 1 0 0 0 is 6 4 1 .