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

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SVP
Joined: 04 May 2006
Posts: 1563
Schools: CBS, Kellogg
What is 1! + 2! + ... + 10! ?  [#permalink]

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14 May 2008, 02:16
4
41
00:00

Difficulty:

35% (medium)

Question Stats:

68% (01:32) correct 32% (01:30) wrong based on 903 sessions

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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

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Math Expert
Joined: 02 Sep 2009
Posts: 56275

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01 Feb 2011, 09:46
4
14
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.

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.

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Joined: 21 Feb 2008
Posts: 55

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14 May 2008, 02:41
9
7
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!

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Is this okay?
##### General Discussion
Manager
Joined: 27 Jul 2007
Posts: 94

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14 May 2008, 02:57
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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14 May 2008, 05:31
2
1
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!

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..
Manager
Joined: 18 Aug 2010
Posts: 74

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01 Feb 2011, 08:09
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 ???
Manager
Joined: 18 Aug 2010
Posts: 74

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01 Feb 2011, 10:12
1
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 !!!
+ KUDO
Manager
Joined: 20 Dec 2013
Posts: 225
Location: India
Re: What is 1! + 2! + ... + 10! ?  [#permalink]

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03 Feb 2014, 08:53
1
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?)

Posted from my mobile device
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Joined: 03 Aug 2015
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Schools: ISB '18, SPJ GMBA '17
GMAT 1: 680 Q48 V35
What is 1! + 2! + ... + 10! ?  [#permalink]

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23 Jan 2016, 01:24
1
Bunuel wrote:
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.

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.

Hello Bunuel,

so in that case, 1!+2!.....1500! will also have the unit digit as 3 only.....

I can remember this concept and apply to any similar questions

Thanks,
Arun
Director
Joined: 23 Jan 2013
Posts: 543
Schools: Cambridge'16
Re: What is 1! + 2! + ... + 10! ?  [#permalink]

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05 Feb 2016, 01:34
can reduce to B and C because sum of numbers is odd. I took C and made mistake
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Re: What is 1! + 2! + ... + 10! ?  [#permalink]

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05 Feb 2016, 07:40
sondenso wrote:
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

CRITICAL OBSERVATION : All options have different Unit digit so calculating unit digit of the sum above will suffice

Unit digit of (1! + 2! + 3! + 4! + 5! + 6!+... + 10!) = 1 + 2+ 6 + 4 + 0 + 0 + 0 + 0+ 0 + 0 = Unit digit 3, Hence

every Factorial of an integer greater than 4 will have Unit digit 0 as it will be a multiple of 10

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Re: What is 1! + 2! + ... + 10! ?  [#permalink]

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16 Aug 2017, 09:48
1!=1
2!=2
3!=6
4!=4 addition of unit's place of these should be sufficient to identify the answer, since beginning 5! we will have 0 for all the other terms.
Option B.
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WE: Engineering (Transportation)
Re: What is 1! + 2! + ... + 10! ?  [#permalink]

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17 Aug 2017, 10:46
Unit digit for
1!=1
2!=2
3!=6
4!=4
and from 5! to 10! units digit is 0
adding them all we get unit digit as 3.

therefore B is only option with unit digit as 3.

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Re: What is 1! + 2! + ... + 10! ?  [#permalink]

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22 Aug 2017, 16:47
sondenso wrote:
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

Notice that the units digit of each answer choice is different; rather than perform the arithmetic as indicated, let’s determine the units digit of each factorial:

1! = 1

2! = 2

3! = 6

4! = 24

After 4!, all other factorials end in zero.

So, the units digit of the answer is 1 + 2 + 6 + 4 = 13, or 3.

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Re: What is 1! + 2! + ... + 10! ?  [#permalink]

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29 Aug 2018, 00:29
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Re: What is 1! + 2! + ... + 10! ?   [#permalink] 29 Aug 2018, 00:29
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