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Source: — Data Sufficiency |
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by neelgandham » Mon Jun 11, 2012 1:50 am
If N is a positive integer, what is the last digit of the sum of the series :1! + 2! + ...N!(1 and N inclusive)?

1! + 2! + ...N!
= 1 + 2 + 6 + 24 + 120 + 720 + ......(From 5! the units digit of all the factorials is 0).
So, if the value of N is greater than or equal to 4 then unit's digit of sum = 1 + 2 + 6 + 24 = 3
1) N is divisible by 4
If N = 4, then unit's digit of sum of the series = 1 + 2 + 6 + 24 = 3
If N = 8, then unit's digit of sum of the series = 1 + 2 + 6 + 24 + 120 + 720 + YX0 + KP0 = 3.
You will get the same result for any other number. So, Statement 1 is sufficient to answer the question.
2) (N^2 + 1)/5 is an Integer
If (N^2 + 1)/5 = 1, N = 2, then unit's digit of sum of the series = 1 + 2 = 3
If (N^2 + 1)/5 = 2, N = 3, then unit's digit of sum of the series = 1 + 2 + 6 = 9
Two different answers. So statement 2 is insufficient to answer the question.


IMO A
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by Ashujain » Mon Jun 11, 2012 5:35 am
St1 is sufficient as explained by neelgandham.

But st2 is also sufficient. below is my explanation:
If (N^2+1)/5 = 1 ---> N=2. Hence last last digit will be 3
Now none of the value of N less than 4 (except 2) satisfies statement2 and hence N = 2 or greater than 4 and in both the cases last digit will be 3.

IMO: D
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by eagleeye » Mon Jun 11, 2012 9:51 am
I agree with Ashujain:
There is no doubt regarding 1) being sufficient, so I will skip it.

I will provide ab it of a mathematical explanation using the fact that (N^2+1)/5 is odd.

We know that N is a positive integer, and that (N^2+1)/5 is odd so let (N^2+1)/5 = 2m+1 (where m is an integer >=0).

Then (N^2+1)/5 =2m+1 => N^2 +1 = 10m+5 => N^2=10m + 4.

Now N is a positive integer (which means N can't be 0), so N^2 is an integer square. The only way this works is if:
N^2=4; => N=2 for which last digit is 3. Now N can be (14,24,34,44,54,64,74 etc.) according to the equation above. But N^2 has to be a perfect square, so for the second smallest value of N,
N^2=64. => N=6.

For the second smallest value of N^2, N=6. Clearly if N>=4, we get the same condition as from statement one where all factorials greater than 4! give 0 as the units digit. So the last digit
of the sum is still 3. SUFFICIENT.

Hence D.

Let me know if this helps :)
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