
#1
Dec106, 07:31 PM

P: 94

Here is a sum from MATHCOUNTS:
What are the last two digits in the sum of the factorials of the first 100 positive integers? From 1! to 4! you can add the units digits, since 5! to ... have 0 in their units place. From that I get 13, and I carry over the 1 over to the next column and add the tens digits of 1! to 9! since 10! to ... have 0 in their tens and units place. I got 73, but the answer key says 13. Can someone please help me? thx 



#2
Dec106, 07:54 PM

P: 1,239

So [tex] \sum_{n=1}^{100} n! [/tex]
[tex] 0!: 01 [/tex] [tex] 1! : 01 [/tex] [tex] 1!+2!: 03[/tex] [tex] 1! + 2! + 3!: 09[/tex] [tex] 1!+2!+3!+4!: 33 [/tex] [tex] 1!+2!+3!+4!+5!: 53 [/tex] Can you see a pattern? Go up to [tex] 9! [/tex] because [tex] \sum_{n=1}^{10} n! [/tex] has the last two digits [tex] 00 [/tex]. Therefore [tex] \sum_{n=1}^{\19} n! [/tex] has last two digits [tex] 13 [/tex] as does [tex] \sum_{n=1}^{100} n! [/tex] 



#3
Dec206, 06:35 AM

Math
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#4
Dec206, 07:52 AM

P: 685

Factorial Sum
1! = 1
2! = 2 3! = 6 4! = 24 5! = 120 6! = 720 7! = 5040 8! = ...20 (the dots stand for some digits, but I didn't calculate them, since I am only interested in the last two digits) 9! = ...80 10! = ...800 Which numbers in the sum of factorials contribute to the last digit? It's 1! = 1 2! = 2 3! = 6 4! = 24 the other numbers have 0 as their last digit. Thus, the last digit of our factorial sum is 3 because 1+2+6+24=33 Which numbers in the sum of factorials contribute to the "10" digit (the digit left to the last digit)? It's 5! = 120 6! = 720 7! = 5040 8! = ...20 9! = ...80 AND the 33 (= sum from 1! to 4!) I wrote down the relevant numbers again and behind the numbers their "10" digit in brackets: 5! = 120 (2) 6! = 720 (2) 7! = 5040 (4) 8! = ...20 (2) 9! = ...80 (8) 33 (3) Let us add the numbers in the brackets: 2+2+4+2+8+3 = 21 Thus, for you sum of factorials the "10" digit is: 1 



#5
Dec206, 09:13 PM

P: 10

Best wishes X = 7 



#6
Dec406, 10:48 AM

P: 1,017

I know this sounds dumb, but Im sorry, I dont see the pattern...




#7
Dec406, 09:00 PM

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#8
Dec506, 09:56 PM

P: 94

Sorry guys... I reworked it and found out I made a mistake... The answer was 71 (taking the last 2 digits of the sum).




#9
Aug210, 01:53 AM

P: 91

It's natural that the last few digits of of the sum of the factorials 1!+2!+3!+..should be the same. The succeeding terms of the series have all zeros (the number of zeroes depending on how many times 2 and 5 appear in the prime factorization of n!) and so do not affect the first digits of the sums representation.




#10
Aug1710, 10:11 AM

P: 1

1!+2!+3!......50!=
wats the answer guys help me out..... 



#11
Aug1710, 06:08 PM

P: 91

31035053229546199656252032972759319953190362094566672920420940313




#12
Aug1810, 01:10 AM

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P: 3,436

lol ripped




#13
Aug1810, 01:12 AM

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#14
Aug1810, 01:30 AM

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P: 3,436

At least he can use the search button! So they kind of cancel each other out in a way.




#15
Aug1910, 03:19 PM

P: 91





#16
Aug2010, 01:47 AM

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#17
Mar1012, 02:42 AM

P: 3

How to find out number of digits in 1!+2!+3!........+100!?




#18
Mar1012, 02:52 AM

P: 4,570

For this problem in base b you need to calculate log_b(z) = ln(z)/ln(b) where b is the number of possibilities in each digit. Round up if you have a fractional part in your answer. The z is your expression 1!+ 2! + 3! + ... blah What you should realize is that for this problem you only need to evaluate the highest term which is 100!. Using properties of logs you can use the property log(ab) = log(a) + log(b) which means the answer for this problem is: log(1) + log(2) + .... log(100) from 1 up to 100 like the sum suggests. 


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