vscid 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

the last digit of factorial of any N>=5 is 0

from statement 1 we get N>=4 hence the last digit will be the last digit of 1!+2!+3!+4!=3-->sufficient

from statement 2, N can be 2 or 8 or so on..now if N is 2 last digit will be 3 and if N is 8 or greater last digit will again be 3
-->this is also sufficient
Ans D either of the statement sufficient

good Q. Thanks for explaination

liferocks wrote:

vscid 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

the last digit of factorial of any N>=5 is 0

from statement 1 we get N>=4 hence the last digit will be the last digit of 1!+2!+3!+4!=3-->sufficient

from statement 2, N can be 2 or 8 or so on..now if N is 2 last digit will be 3 and if N is 8 or greater last digit will again be 3
-->this is also sufficient
Ans D either of the statement sufficient

1 if N is divisible by 4 it means it may be 4 , 8,12 ....so on we cant conclude anything ----- so insufficient alone.
2. N^2+1/5 is an odd integer

if N = 8 => 64+1/5= 13 and if N = 12 then 144+1/5 = 29 is odd SO we can not find the exactly what N is 8 or 12 so how can we get the value of 1!+2!+3!+....N! so 2 is also not sufficient alone..


so both 1 and 2 together not sufficient

can any expert help me?

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

1! = 1
1! + 2! = 3
1! + 2! + 3! = 9
1! + 2! + 3! + 4! = 33
1! + 2! + 3! + 4! + 5! = 153
1! + 2! + 3! + 4! + 5! + 6! = 873
----
so on ------- Last digit would be 3

Statement 1:
N = 4, 8, 12 etc
Unit digit = 3
Sufficient

Statement 2:
(N^2 + 1) / 5 = odd integer
N^2 + 1 = 5, 15, 25, 35, ..., 65, ...
N^2 = 4, 14, 24, 34, ..., 64, ...
N = 2, 8, ...
Unit Digit = 3
Sufficient

Answer: D