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# M15-15

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Math Expert
Joined: 02 Sep 2009
Posts: 47898

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16 Sep 2014, 00:55
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Difficulty:

95% (hard)

Question Stats:

41% (02:12) correct 59% (01:26) wrong based on 124 sessions

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If $$N$$ is a positive integer, what is the units digit of $$1! + 2! + ... +N!$$?

(1) $$N$$ is divisible by 4

(2) $$\frac{N^2 + 1}{5}$$ is an odd integer

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Joined: 02 Sep 2009
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16 Sep 2014, 00:55
3
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Official Solution:

Generally the units digit of $$1! + 2! + ... +N!$$ can take ONLY 3 values:

A. If $$N=1$$ then the units digit is 1;

B. If $$N=3$$ then the units digit is 9;

C. If $$N= \{ \text{ANY other value} \}$$, then the units digit is 3 (if $$N=2$$, then $$1!+2!=3$$; if $$N=4$$, then $$1!+2!+3!+4!=33$$ and if $$N \ge 4$$, then the terms after $$N=4$$ will end by 0 thus will not affect the units digit and it'll remain 3).

So basically question asks whether we can determine which of the three cases we have.

(1) $$N$$ is divisible by 4. This statement implies that $$N$$ is not 1 or 3, thus third case. Sufficient.

(2) $$\frac{N^2 + 1}{5}$$ is an odd integer. The same here: $$N$$ is not 1 or 3, thus third case. Sufficient.

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Joined: 21 May 2014
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WE: Account Management (Manufacturing)

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26 Oct 2014, 11:24
Bunuel wrote:
Official Solution:

Generally the units digit of $$1! + 2! + ... +N!$$ can take ONLY 3 values:

A. If $$N=1$$ then the units digit is 1;

B. If $$N=3$$ then the units digit is 9;

C. If $$N= \{ \text{ANY other value} \}$$, then the units digit is 3 (if $$N=2$$, then $$1!+2!=3$$; if $$N=4$$, then $$1!+2!+3!+4!=33$$ and if $$N \ge 4$$, then the terms after $$N=4$$ will end by 0 thus will not affect the units digit and it'll remain 3).

So basically question asks whether we can determine which of the three cases we have.

(1) $$N$$ is divisible by 4. This statement implies that $$N$$ is not 1 or 3, thus third case. Sufficient.

(2) $$\frac{N^2 + 1}{5}$$ is an odd integer. The same here: $$N$$ is not 1 or 3, thus third case. Sufficient.

For Statement 2, you can also take N=2 which will give (4+1)/5=1 odd integer .. now unit digit will be 3 because of N=2
if take take N=12 again answer will be odd and this time unit digit will be 0
This makes statement 2 insufficient right ?
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Joined: 02 Sep 2009
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27 Oct 2014, 06:29
2
awal_786@hotmail.com wrote:
Bunuel wrote:
Official Solution:

Generally the units digit of $$1! + 2! + ... +N!$$ can take ONLY 3 values:

A. If $$N=1$$ then the units digit is 1;

B. If $$N=3$$ then the units digit is 9;

C. If $$N= \{ \text{ANY other value} \}$$, then the units digit is 3 (if $$N=2$$, then $$1!+2!=3$$; if $$N=4$$, then $$1!+2!+3!+4!=33$$ and if $$N \ge 4$$, then the terms after $$N=4$$ will end by 0 thus will not affect the units digit and it'll remain 3).

So basically question asks whether we can determine which of the three cases we have.

(1) $$N$$ is divisible by 4. This statement implies that $$N$$ is not 1 or 3, thus third case. Sufficient.

(2) $$\frac{N^2 + 1}{5}$$ is an odd integer. The same here: $$N$$ is not 1 or 3, thus third case. Sufficient.

For Statement 2, you can also take N=2 which will give (4+1)/5=1 odd integer .. now unit digit will be 3 because of N=2
if take take N=12 again answer will be odd and this time unit digit will be 0
This makes statement 2 insufficient right ?

Not right.

If N is NOT 1 or 3, then the units digit of $$1! + 2! + ... +N!$$ is 3.
If N = 2, the units digit is 3 --> $$1! + 2!=3$$.
If N = 12, the unit digits is 3 too --> $$1! + 2! + ... +12!=522,956,313$$
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26 Jun 2016, 04:45
I think this is a poor-quality question and I agree with explanation. Do you think GMAT will expect us to know this property?
Intern
Joined: 21 May 2016
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19 Nov 2016, 22:24
But isnt 1!+2!+...N! where N=1 equal to 4?
1+2+1 = 4
Also when N=2, it becomes 1!+2!+2! = 1+2+2=5, why are we saying that the units digit is 3?
Please let me know where I'm going wrong
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20 Nov 2016, 03:35
Bhavanahiremath wrote:
But isnt 1!+2!+...N! where N=1 equal to 4?
1+2+1 = 4
Also when N=2, it becomes 1!+2!+2! = 1+2+2=5, why are we saying that the units digit is 3?
Please let me know where I'm going wrong

If N = 1, we have only 1! = 1
If N = 2, then we have 1! + 2! = 3
If N = 3, then we have 1! + 2! + 3! = 9
...
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02 Jan 2017, 05:42
Is this actually needed for GMAT?!
Intern
Joined: 27 Oct 2015
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03 Jan 2017, 06:44
Hi Buenel,

Is it not true that even 1 is divisible by 4?

Also (1 square +4)/5 is 1 , which is an odd integer.

Thus, 1 satisfies both the conditions and thus I believe that the correct answer choice should be E.

Where am I erring?

Many Thanks.
Math Expert
Joined: 02 Sep 2009
Posts: 47898

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04 Jan 2017, 01:16
dsheth7 wrote:
Hi Buenel,

Is it not true that even 1 is divisible by 4?

Also (1 square +4)/5 is 1 , which is an odd integer.

Thus, 1 satisfies both the conditions and thus I believe that the correct answer choice should be E.

Where am I erring?

Many Thanks.

Integer x is divisible by integer y means that x/y = integer. 1/4 = 0.25, which is not an integer. Thus 1 is not divisible by 4.
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05 Jan 2017, 07:09
I think another way to solve this problem would be using the rule below:

Any multiplication series which contains 2*5 will always have units digit as zero.

So just calculate values till 4! after which the unit digit should be the same, as from n=5, all terms from 5! will have units as zero.

(n=2) => 1+ 2!=3
(n=3)=> 3+ 3! = 9
(n=4)=> 9+ 4! = 33
After this all the terms that you add will have units digit as zero as they contain 2*5 so units digit is always 3.
(n=5) =>33+5! (contains 2*5) =123
(n=5) =>33+ 6! (contains 2*5)= 753 and so on.
Hence, both statements are sufficient.

Thanks Bunuel, brilliant question!
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Joined: 26 Jun 2014
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06 Feb 2017, 11:02
Bhavanahiremath wrote:
But isnt 1!+2!+...N! where N=1 equal to 4?
1+2+1 = 4
Also when N=2, it becomes 1!+2!+2! = 1+2+2=5, why are we saying that the units digit is 3?
Please let me know where I'm going wrong

Is it wrong? I didnt find any.
Intern
Joined: 25 Sep 2017
Posts: 1

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04 Oct 2017, 05:39
This was my reasoning for S2 without knowing the general rule about the unit digits you wrote about, I found the right answer but please tell me if the reasoning is flawed:

if (N(^2)+1) / 5 is an odd integer, then the nominator ( N^2 + 1 ) must be an odd multiple of 5 ( 15, 25, 35 ...). We know that N is an integer, so that odd muliple of 5 -1 must be an even perfect square square such as: N^2 = (odd multiple of 5) - 1= perfect even perfect square. From that : N^2 = 65 - 1 = 64 , N=8, plug it in the set: it ends with a 3.
Manager
Joined: 09 Feb 2018
Posts: 66
GMAT Date: 03-15-2018

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17 Jun 2018, 12:45
Really good question . Loved it
Intern
Joined: 11 Jun 2018
Posts: 1

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24 Jul 2018, 04:55
Hi,

What if N=4? 1!+2!+3!+4!= 33.
Math Expert
Joined: 02 Sep 2009
Posts: 47898

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24 Jul 2018, 05:01
kartmn wrote:
Hi,

What if N=4? 1!+2!+3!+4!= 33.

What is your question? Isn't the units digit 3? We got from each of the statements that the units digit must be 3
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Re: M15-15 &nbs [#permalink] 24 Jul 2018, 05:01
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