Last Digit of factors

[Uva@90]

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.

[theCodeToGMAT]

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"
above this all will have "3" as unit digit
SUFFICIENT

Answer [spoiler]{D}[/spoiler]

[Uva@90]

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]

Thanks Rahul,
I got your solution.
Take away from here any factorial(integer) greater than 4 end with 0.

Regards,
Uva.

[sanju09]

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.

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.

1! + 2! = 1 + 2 = 3.

Hence [spoiler]D[/spoiler].

[ankitbagla]

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.

[Brent@GMATPrepNow]

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.

Let's take theCodeToGMAT's great solution and examine a few more terms.

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
.
.
.

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.

Statement 1 ensures that N > 4

Statement 2 ensures that N is either 2 or an integer greater than 4. In both of these cases, the units digit is guaranteed to be 3.

Cheers,
Brent

[sanju09]

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.

Let's take theCodeToGMAT's great solution and examine a few more terms.

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
.
.
.

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

I don't think that St (2) ensures that N ≥ 4, it's possible at N = 2 too.

[Brent@GMATPrepNow]

I don't think that St (2) ensures that N ≥ 4, it's possible at N = 2 too.

Good catch - I edited my response accordingly.

Cheers,
Brent

[sanju09]

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]

In fact most of the positive integers N ending in 2, 3, 7, or 8 are fit in St (2). Do you agree Brent?

No, I don't agree.

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

[kop]

In fact most of the positive integers N ending in 2, 3, 7, or 8 are fit in St (2). Do you agree Brent?

No, I don't agree.

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 Gurus,
Please explain me the stat2 more elaborately:
Suppose n=12 and (N² + 1)/5 = 145/5 = 29 . In this case unit digit is not 3??
Please explain.
Thanks

[kop]

In fact most of the positive integers N ending in 2, 3, 7, or 8 are fit in St (2). Do you agree Brent?

No, I don't agree.

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 Gurus,
Please explain me the stat2 more elaborately:
Suppose n=12 and (N² + 1)/5 = 145/5 = 29 . In this case unit digit is not 3??
Please explain.
Thanks

[theCodeToGMAT]

Kop, in my solution we have observed a pattern that starting from N=5 all the digits will have unit digit 0 because factors 2 & 5 will be present in all those factorials.

According to statement 2, N can be 2, 4, etc.. Because they will yield odd quotient .

Now, 1! +2!= 3
And, 1!+2!+3!+4!= 33
Both have 3 as unit digit.

Also, as we stated, 5! Onwards wil have unit digit as 0.. So, unit digit will be "3" only

[mevicks]

Suppose n=12 and (N² + 1)/5 = 145/5 = 29 . In this case unit digit is not 3??

Hi Kop,

You are losing sight of the target (original) question this DS asks:
Q: what is the last digit of 1! + 2! + ... +N!

You have correctly pointed out that when n = 12 the statement 2 yields a value 29 (which is odd) BUT we need to go a step further and evaluate the TARGET question with this value of n (although not required as we have crossed the problematic barrier of n = 4)

When n = 12 (i.e greater than 4) the sum of the first 12 factorials would always yield a units digit 3 (dont forget that n = 2 also yields a units digit 3)

Thus st2. is SUFFICIENT

Hope that helps.

[sanju09]

In fact most of the positive integers N ending in 2, 3, 7, or 8 are fit in St (2). Do you agree Brent?

No, I don't agree.

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

Oh yeah! I involved both even and odd in my generalization and that again is incomplete. I should have meant that

In fact all the positive integers N ending in 2,3, 7, or 8 yield a positive integer in St (2).

But we've to look for the odd integers only, which is obtained only when the positive integer N ends in a 2 or 8. You caught me right Brent, score settled