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# m12 Q 25

Author Message
Intern
Status: GMAT once done, Going for GMAT 2nd time.
Joined: 29 Mar 2010
Posts: 34
Location: Los Angeles, CA
Schools: Anderson, Haas, Ross, Kellog, Booth, McCombs
Followers: 0

Kudos [?]: 3 [0], given: 3

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05 Jun 2011, 20:01
What is 1! + 2! + ... + 10!?

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

What is the easy way to solve this question? Thanks in advance.
_________________

BP

Senior Manager
Joined: 24 Mar 2011
Posts: 450
Location: Texas
Followers: 5

Kudos [?]: 184 [0], given: 20

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05 Jun 2011, 21:24
I won't solve it, but thinking it logically to choose the correct answer.

you know that there is 1! so final addition has to be odd.
So you can knock out 1st, 4th and 5th option.

Now there is little process to pick one of the 2nd or 3rd.

excluding 1! you have, 2!+3!+.....+9!+10!
to pick one choice some of these numbers has to be either 4037912 or 4037914
now see the number of 2's in above series..

2 + 2^1 *3 + 2^3*3 + 2^3 *3*5 + ....

= 2(1 + 3 + 2^2*3 + ..)

if you divide above numbers by 2 both are divisible, so you need to go further -

= 2(4 + 2^2 * 3 +... )
= 2*2^2 (1 + 3 +..)
now if you notice 4037912 is divided by 8 but not 4037917.

Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2013
Followers: 163

Kudos [?]: 1825 [2] , given: 376

### Show Tags

06 Jun 2011, 02:36
2
KUDOS
valencia wrote:
What is 1! + 2! + ... + 10!?

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

What is the easy way to solve this question? Thanks in advance.

Unit's digit of 1!+2!+3!+4!=3
Unit's digit of the sum of the remaining integers will always be 0 i.e. 5!+6!+.....+$$\infty !$$=0
Thus, the unit's digit of the above summation=3+0=3
Only B fits.

Ans: "B"
_________________
Re: m12 Q 25   [#permalink] 06 Jun 2011, 02:36
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# m12 Q 25

Moderator: Bunuel

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