IMO D

1) N is divisible by 4

If N=4 , 1!+2!+3!+4! = 1+2+6+24 =33, so Last digit is 3
If N=8 , 1!+2!+3!+4!+......+ 8! =??

Now we know 5!= 120 and further multiplication with any integer (6,7,8 etc) will result in 0 as its unit's digit.
so 1!+2!+3!+4!+......+ 8! = 33+ 120+ ..0+...0+ ....0 = ......3

So last digit is 3 , Sufficient

2) (N^2+1)/5 is an odd integer

Both N= 8 and N= 18 satisfy statement 2 .

Now substituting each values in 1!+2!+3!+4!...+ N! will give the last digits as 3 in both the cases (same logic as mentioned above).

So last digit is 3 , Sufficient

pankajks2010 wrote:If N is a positive integer, what is the last digit of the result for 1!+2!+...N!?

a) N is divisible by 4
b) (N^2+1)/5 is an odd integer

Courtesy: 800score.com
Not sure, if this question is already discussed on the forum or not.

1!+2!+......N!

1) N is divisible by 4;
N=4k; if N=4; 1+2!+3!+4!=1+2+6+24=33; last digit is 3;
now consider analyze this, 5! onwards every factorial is going to end with 0, because it contains 5, hence 1+2!+3!+4!+5!+6!+7!+8!; will also have 3 as its last digit, hence 1 alone is sufficient to answer the question.

2)(N^2+1)/5 is an odd integer;
i.e. last digit of N^2+1 must be 5 hence last digit of N^2 must be 4; now possible no. here would be 8!,12!,18!,22!,28!.....;

again by same reasoning, when we add these no. they will always have 3 as its last digit; hence 2 alone is also sufficient to answer the question, hence answer should be D

why should we make assumption of consecutive integers (sequence) when it says

If N is a positive integer, what is the last digit of the result for 1!+2!+...N!?

it wouldn't mention N as +ve in case of consecutive no-s

isn't it c

what's OA? (explanation after)

Manpsingh got it spot on!!..The OA is D.

800score.com is an awesome source, but sometimes their wording is guess-through unlike that of GMAT official source

manpsingh87: again by same reasoning, when we add these no. they will always have 3 as its last digit; hence 2 alone is also sufficient to answer the question, hence answer should be D

Thanks manpsingh87

I forgot to apply values of N(8,18) to.. 1!+2!+....N! to get the last digits :(
(These are some of the common silly mistakes I have been doing now and then. I hope my mind functions right in the real exam! )

Definitely D is the answer.

If N is a positive integer, what is the last digit of the result for 1!+2!+...N!?

a) N is divisible by 4
b) (N^2+1)/5 is an odd integer


This question is such that even if you do not give any of the statements, and only say that N is a positive integer (which is given), still the last digit would end in 3.

Thats the best part of this question...