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m12 Q 25

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Status: GMAT once done, Going for GMAT 2nd time.
Joined: 29 Mar 2010
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Location: Los Angeles, CA
Schools: Anderson, Haas, Ross, Kellog, Booth, McCombs
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m12 Q 25 [#permalink] New post 05 Jun 2011, 19:01
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.
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Re: m12 Q 25 [#permalink] New post 05 Jun 2011, 20:24
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.

So B is the answer.
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Re: m12 Q 25 [#permalink] New post 06 Jun 2011, 01:36
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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"
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Re: m12 Q 25   [#permalink] 06 Jun 2011, 01:36
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