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What is 1! + 2! + ... + 10! ?

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What is 1! + 2! + ... + 10! ? [#permalink] New post 14 May 2008, 01:16
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A
B
C
D
E

Difficulty:

  25% (low)

Question Stats:

68% (02:10) correct 31% (01:18) wrong based on 97 sessions
What is 1! + 2! + ... + 10! ?

A. 4,037,910
B. 4,037,913
C. 4,037,915
D. 4,037,916
E. 4,037,918
[Reveal] Spoiler: OA

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Re: M12-25 [#permalink] New post 14 May 2008, 01:41
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sondenso wrote:
What is 1!+2!+...+10! ?


4,037,910
4,037,913
4,037,915
4,037,916
4,037,918

I think that if I meet this on the real Gmat, I must waste 10 minute, even 1 hour 15 minutes more!


There's a easier way to crack this...

1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
Now, all figures after 5! will have 0 in their unit digit (720 on 6!, 5040 on 7! etc.) so, all that we have to look into is the unit's place in the answer!

1+2+6+4 = 13 so, the answer should end with 3. Pick your answer!
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Re: M12-25 [#permalink] New post 14 May 2008, 01:57
B.

1+2*1+3*2*1+..................
= 1+2[1+3[1+4[1+5[1+6[.................
unit's digit => 3
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Re: M12-25 [#permalink] New post 14 May 2008, 04:31
sondenso wrote:
What is 1!+2!+...+10! ?


4,037,910
4,037,913
4,037,915
4,037,916
4,037,918

I think that if I meet this on the real Gmat, I must waste 10 minute, even 1 hour 15 minutes more!



how about 15 seconds??? answer is B
just look at the unit digit
1!=1
2!=2
3!=6
4!=4
5!=0..after this all numbers will have a 2 and 5..and thus unit digit for all of them is 0..

1+2+6+4=13..luckily 13, 3 is only one of the ans choices..
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Re: M12-25 [#permalink] New post 01 Feb 2011, 07:09
hello, Can somebody explain /explain whats wrong with my method:

[(1+10!)/2] *10 ( average * number of terms)

i am getting 5 as last unit digit ???
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Re: M12-25 [#permalink] New post 01 Feb 2011, 08:46
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tinki wrote:
hello, Can somebody explain /explain whats wrong with my method:

[(1+10!)/2] *10 ( average * number of terms)

i am getting 5 as last unit digit ???


You can apply the formula above: Sum=\frac{first+last}{2}*# \ of \ terms, the mean multiplied by the number of terms for evenly spaced set (aka arithmetic progression). But as in the sequence given (1!, 2!, 3!, ..., 10!) the difference between any two successive terms is not the same then we don't have evenly spaced set (arithmetic progression) and thus can not apply this formula.

Check this thread for more: math-number-theory-88376.html also this thread: sequences-progressions-101891.html

As for the solution:
What is 1!+2!+...+10! ?
A. 4,037,910
B. 4,037,913
C. 4,037,915
D. 4,037,916
E. 4,037,918

For any integer n more than or equal to 5 the units digit of n! will be zero as n!, in this case, will contain at least one 2 and 5 which when multiplied will give a trailing zero. So terms from 5! to 10! will have zero as their units digit. 1!+2!+3!+4!=1+2+6+24=33 so the whole sum will have 3 as the units digit. Only option B offers a number with 3 as its units digit.

Answer: B.
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Re: M12-25 [#permalink] New post 01 Feb 2011, 09:12
got it. somehow thought factorials were evenly spaced. silly me ...


i saw the material. ARE you kind of magician ?

you are like : "you need help? Here i am" THANKS SOOOO MUCH, YOU ARE DEFINITELY GREAT !!! :)
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Re: What is 1! + 2! + ... + 10! ? [#permalink] New post 03 Feb 2014, 05:27
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Re: What is 1! + 2! + ... + 10! ? [#permalink] New post 03 Feb 2014, 07:53
Taking units digits of 1! To 4! ie 1+2+6+4+0=13(Because only the units digits are diff in options).After that 0 will be at the unit's place for 5!,6!,7!,8!,9!,10!.Therefore,the last two digits will be 13.
Ans.B

(Is the logic above flawed?)

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Re: What is 1! + 2! + ... + 10! ?   [#permalink] 03 Feb 2014, 07:53
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