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M12-25

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Math Expert
Joined: 02 Sep 2009
Posts: 46297

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16 Sep 2014, 00:47
1
8
00:00

Difficulty:

25% (medium)

Question Stats:

70% (00:46) correct 30% (00:36) wrong based on 124 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

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 46297

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16 Sep 2014, 00:47
Official 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

$$1! + 2! + ... + 10! = 1 + (2! + ... + 10!)$$ = odd + even = odd.

$$1! + 2! + ... + 10! = 3 + (3! + ... + 10!)$$ = integer divisible by 3.

Among the listed choices we are looking for odd integers divisible by 3. Only choice B fits.

_________________
Intern
Joined: 22 Jul 2013
Posts: 19
Location: United States
Concentration: Technology, Entrepreneurship
Schools: IIM A '15
GMAT 1: 650 Q46 V34
GMAT 2: 720 Q49 V38
GPA: 3.67
WE: Engineering (Non-Profit and Government)

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22 Oct 2014, 18:17
6
3
just check units digit of the sum

Units digit
1!=1
2!=2
3!=6
4!=4
5! and above 0

so 6+4+2+1 = 13 or units digit 3. Only answer choice that ends with 3 is B:-)

Bunuel wrote:
Official 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

$$1! + 2! + ... + 10! = 1 + (2! + ... + 10!)$$ = odd + even = odd.

$$1! + 2! + ... + 10! = 3 + (3! + ... + 10!)$$ = integer divisible by 3.

Among the listed choices we are looking for odd integers divisible by 3. Only choice B fits.

Board of Directors
Joined: 17 Jul 2014
Posts: 2730
Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30
GPA: 3.92
WE: General Management (Transportation)

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22 Nov 2014, 20:58
amariappan wrote:
just check units digit of the sum

Units digit
1!=1
2!=2
3!=6
4!=4
5! and above 0

so 6+4+2+1 = 13 or units digit 3. Only answer choice that ends with 3 is B:-)

Bunuel wrote:
Official 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

$$1! + 2! + ... + 10! = 1 + (2! + ... + 10!)$$ = odd + even = odd.

$$1! + 2! + ... + 10! = 3 + (3! + ... + 10!)$$ = integer divisible by 3.

Among the listed choices we are looking for odd integers divisible by 3. Only choice B fits.

that is exactly how I got the answer
Intern
Joined: 22 Jul 2013
Posts: 19
Location: United States
Concentration: Technology, Entrepreneurship
Schools: IIM A '15
GMAT 1: 650 Q46 V34
GMAT 2: 720 Q49 V38
GPA: 3.67
WE: Engineering (Non-Profit and Government)

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22 Nov 2014, 22:10
[quote=mvictor]
great thinking!!! All the best my friend.

digit
1!=1
2!=2
3!=6
4!=4
5! and above 0

so 6+4+2+1 = 13 or units digit 3. Only answer choice that ends with 3 is B:-)

Bunuel wrote:
Official 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

$$1! + 2! + ... + 10! = 1 + (2! + ... + 10!)$$ = odd + even = odd.

$$1! + 2! + ... + 10! = 3 + (3! + ... + 10!)$$ = integer divisible by 3.

Among the listed choices we are looking for odd integers divisible by 3. Only choice B fits.

[/quote]

that is exactly how I got the answer [/quote]
Intern
Joined: 14 Oct 2015
Posts: 36
GMAT 1: 640 Q45 V33

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05 Nov 2015, 06:53
mvictor wrote:
amariappan wrote:
just check units digit of the sum

Units digit
1!=1
2!=2
3!=6
4!=4
5! and above 0

so 6+4+2+1 = 13 or units digit 3. Only answer choice that ends with 3 is B:-)

Bunuel wrote:
Official 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

$$1! + 2! + ... + 10! = 1 + (2! + ... + 10!)$$ = odd + even = odd.

$$1! + 2! + ... + 10! = 3 + (3! + ... + 10!)$$ = integer divisible by 3.

Among the listed choices we are looking for odd integers divisible by 3. Only choice B fits.

that is exactly how I got the answer

I agree. This may even be faster than the number properties approach. Maybe 2 or 3 of the answer choices could have 3 as unit digit to make it tricky.
Verbal Forum Moderator
Joined: 15 Apr 2013
Posts: 190
Location: India
Concentration: General Management, Marketing
GMAT Date: 11-23-2015
GPA: 3.6
WE: Science (Other)

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15 Nov 2015, 10:39
5
A useful property of factorials is that consecutive factorials (in increasing order) are multiples of previous factorials.

1!

2! = 1! X 2

3!= 2! X 3

4!= 3! X 4

5!= 4! x 5

Secondly all the factorials greater than/equal to 5 (eg. 5!, 6!, 7!, 8! would end in zero because each would have primes 2 and well as 5)

Now back to the question:

1!+2! ........+10!

Take total of first four factorials:

1! = 1

2! = 1! X 2 = 2

3!= 2! X 3 = 6

4!= 3! X 4 = 24

Total is 33

higher factorials would have unit digits as zero. So unit digit of total is 3.
Current Student
Joined: 29 Apr 2015
Posts: 26
Location: Russian Federation
GMAT 1: 710 Q48 V38
GPA: 4

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31 Jan 2016, 02:19
Bunuel 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

I have another approach.
All choices are different from each other only by digit units. So, we have to find the digit unit of the sought quantity.
All factorials since 5! have digit unit "0", because they are consist of factors 5 and 2.
Therefore, we have to count the value of 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33.
So, digit unit of the sought value is 3.
Intern
Joined: 21 Dec 2014
Posts: 31
GMAT 1: 710 Q49 V37

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13 Oct 2016, 22:51
I think this is a high-quality question and I agree with explanation. i have another way to solve this problem.
5!to 10!, the unit digit must be 0.
3!+4!=30(unit digit is also 0)
So, unit digit of 1!+2!....10! must be 1!+2!=3
==> choice B.
Does it make sense.
Manager
Joined: 27 Aug 2014
Posts: 56
Concentration: Strategy, Technology
GMAT 1: 660 Q45 V35
GPA: 3.66
WE: Consulting (Consulting)

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17 Jul 2017, 18:13
1 + 2 + 6 + 24 + 120 + 720 + anything that ends in 0..

so 1+2+6+4 = something that ends in 3.

So thats how I got the answer.
Intern
Joined: 24 Oct 2016
Posts: 29

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19 Sep 2017, 06:36
Can we algebraically solve such questions ?
Re: M12-25   [#permalink] 19 Sep 2017, 06:36
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