theCodeToGMAT wrote: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

sanju09 wrote: Be observant! N cannot take 4 in St (2). All it can take is a positive integer ending in 2 or 8 only.