good DS Qs.
This topic has expert replies
- neelgandham
- Community Manager
- Posts: 1060
- Joined: Fri May 13, 2011 6:46 am
- Location: Utrecht, The Netherlands
- Thanked: 318 times
- Followed by:52 members
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
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.
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
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
If N = 4, then unit's digit of sum of the series = 1 + 2 + 6 + 24 = 31) N is divisible by 4
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.
If (N^2 + 1)/5 = 1, N = 2, then unit's digit of sum of the series = 1 + 2 = 32) (N^2 + 1)/5 is an Integer
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
Anil Gandham
Welcome to BEATtheGMAT | Photography | Getting Started | BTG Community rules | MBA Watch
Check out GMAT Prep Now's online course at https://www.gmatprepnow.com/
Welcome to BEATtheGMAT | Photography | Getting Started | BTG Community rules | MBA Watch
Check out GMAT Prep Now's online course at https://www.gmatprepnow.com/
- Ashujain
- Master | Next Rank: 500 Posts
- Posts: 123
- Joined: Fri Apr 20, 2012 4:36 am
- Thanked: 29 times
- Followed by:1 members
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
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
- eagleeye
- Legendary Member
- Posts: 520
- Joined: Sat Apr 28, 2012 9:12 pm
- Thanked: 339 times
- Followed by:49 members
- GMAT Score:770
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
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