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M12-25

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M12-25 [#permalink]

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New post 16 Sep 2014, 00:47
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A
B
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Difficulty:

  25% (medium)

Question Stats:

71% (00:46) correct 29% (00:35) wrong based on 112 sessions

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Kudos [?]: 128512 [1], given: 12179

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Re M12-25 [#permalink]

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New post 16 Sep 2014, 00:47
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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.


Answer: B
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Re: M12-25 [#permalink]

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New post 22 Oct 2014, 18:17
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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.


Answer: B

Kudos [?]: 20 [6], given: 3

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Re: M12-25 [#permalink]

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New post 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.


Answer: B


that is exactly how I got the answer ;)

Kudos [?]: 392 [0], given: 182

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Re: M12-25 [#permalink]

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New post 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.


Answer: B
[/quote]

that is exactly how I got the answer ;)[/quote]

Kudos [?]: 20 [0], given: 3

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Re: M12-25 [#permalink]

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New post 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.


Answer: B


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.

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M12-25 [#permalink]

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New post 15 Nov 2015, 10:39
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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.

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Re: M12-25 [#permalink]

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New post 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.

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Re M12-25 [#permalink]

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New post 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.

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Re: M12-25 [#permalink]

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New post 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.

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Re: M12-25 [#permalink]

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New post 19 Sep 2017, 06:36
Can we algebraically solve such questions ?

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Re: M12-25   [#permalink] 19 Sep 2017, 06:36
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M12-25

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