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Re M1515 [#permalink]
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16 Sep 2014, 00:55
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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. Answer: D
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Re: M1515 [#permalink]
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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.
Answer: D 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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Re: M1515 [#permalink]
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27 Oct 2014, 06:29
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.
Answer: D 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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Re M1515 [#permalink]
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26 Jun 2016, 04:45
I think this is a poorquality question and I agree with explanation. Do you think GMAT will expect us to know this property?



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Re: M1515 [#permalink]
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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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Re: M1515 [#permalink]
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20 Nov 2016, 03:35



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Re: M1515 [#permalink]
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02 Jan 2017, 05:42
Is this actually needed for GMAT?!



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Re: M1515 [#permalink]
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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.



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Re: M1515 [#permalink]
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04 Jan 2017, 01:16



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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. Please see below for clarity:
(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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Re: M1515 [#permalink]
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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.



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Re: M1515 [#permalink]
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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.










