Thanks Rahul,theCodeToGMAT wrote:N --> Positive
1! = 1
1! + 2! = 3
1! + 2! +3! = 9
1! + 2! +3! + 4! = xx3
From "5!" each term will have "0" as last digit because the factors will consist 2 & 5
Statement 1:
N is divisible by 4
N >= 4
Unit digit "3"
SUFFICIENT
Statement 2:
(N^2 + 1)/5
The number must be divisible by "5"
N = 2 , unit Digit = "3"
N = 4 , Unit Digit = 3
above this all will have "3" as unit digit
SUFFICIENT
Answer [spoiler]{D}[/spoiler]
Nicely attempted by Rahul. Message is this simple: as long as N is 4 or more, the unit's digit of 1! + 2! + ... +N! is 3. The statement (1) confirms that N is 4 or more, and to add little to that since N = 2 too satisfy the St (2), still the answer remains to be 3.Uva@90 wrote:If N is a positive integer, what is the last digit of 1! + 2! + ... +N!?
(1) N is divisible by 4
(2) (N^2 + 1)/5 is an odd integer.
OA D
Regards,
Uva.
Let's take theCodeToGMAT's great solution and examine a few more terms.ankitbagla wrote:as per my understanding
if N is div by 4 then N can be 4,8,12, and so on . But we don't have to look for N>5 because in that case last digit will be 0. but when N=4 then last digit is 5. how can this alone be sufficient?
Please correct me if i am wrong in my approach.
I don't think that St (2) ensures that N ≥ 4, it's possible at N = 2 too.Brent@GMATPrepNow wrote:Let's take theCodeToGMAT's great solution and examine a few more terms.ankitbagla wrote:as per my understanding
if N is div by 4 then N can be 4,8,12, and so on . But we don't have to look for N>5 because in that case last digit will be 0. but when N=4 then last digit is 5. how can this alone be sufficient?
Please correct me if i am wrong in my approach.
1! = 1
1! + 2! = 1 + 2 = 3
1! + 2! +3! = 1 + 2 + 6 = 9
1! + 2! +3! + 4! = 1 + 2 + 6 + 24 = 33
1! + 2! +3! + 4! + 5! = 1 + 2 + 6 + 24 + 120 = 153
1! + 2! +3! + 4! + 5! + 6! = 1 + 2 + 6 + 24 + 120 + 720= 873
1! + 2! +3! + 4! + 5! + 6! + 7! = 1 + 2 + 6 + 24 + 120 + 720 + 5040 = 5913
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.
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As you can see, once we get to adding 5!, EVERY sum will have 3 as its units digit.
So, if N > 4, the sum will have 3 as its units digit.
Statements 1 and 2 each ensure that N > 4
Cheers,
Brent
Good catch - I edited my response accordingly.sanju09 wrote: I don't think that St (2) ensures that N ≥ 4, it's possible at N = 2 too.
No, I don't agree.sanju09 wrote:In fact most of the positive integers N ending in 2, 3, 7, or 8 are fit in St (2). Do you agree Brent?
Brent@GMATPrepNow wrote:No, I don't agree.sanju09 wrote:In fact most of the positive integers N ending in 2, 3, 7, or 8 are fit in St (2). Do you agree Brent?
First of all, I'm not sure how such a generalization would help us.
For statement 2, once we've ruled out the possibility that N = 1 or 3 (and shown that N = 2 yields a units digit of 3) then we have sufficiency (since all values of N > 4 also yield a units digit of 3).
Second, and more importantly, positive integers N ending in 3 and 7 will never meet the condition that (N² + 1)/5 is an ODD integer.
Cheers,
Brent
Brent@GMATPrepNow wrote:No, I don't agree.sanju09 wrote:In fact most of the positive integers N ending in 2, 3, 7, or 8 are fit in St (2). Do you agree Brent?
First of all, I'm not sure how such a generalization would help us.
For statement 2, once we've ruled out the possibility that N = 1 or 3 (and shown that N = 2 yields a units digit of 3) then we have sufficiency (since all values of N > 4 also yield a units digit of 3).
Second, and more importantly, positive integers N ending in 3 and 7 will never meet the condition that (N² + 1)/5 is an ODD integer.
Cheers,
Brent
Hi Kop,kop wrote: Suppose n=12 and (N² + 1)/5 = 145/5 = 29 . In this case unit digit is not 3??
Oh yeah! I involved both even and odd in my generalization and that again is incomplete. I should have meant thatBrent@GMATPrepNow wrote:No, I don't agree.sanju09 wrote:In fact most of the positive integers N ending in 2, 3, 7, or 8 are fit in St (2). Do you agree Brent?
First of all, I'm not sure how such a generalization would help us.
For statement 2, once we've ruled out the possibility that N = 1 or 3 (and shown that N = 2 yields a units digit of 3) then we have sufficiency (since all values of N > 4 also yield a units digit of 3).
Second, and more importantly, positive integers N ending in 3 and 7 will never meet the condition that (N² + 1)/5 is an ODD integer.
Cheers,
Brent
