factorials

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vscid: Master | Next Rank: 500 Posts; Posts: 212; Joined: Sat Dec 01, 2007 4:19 pm; Thanked: 5 times

factorials

by vscid » Fri Apr 16, 2010 5:38 pm

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 GMAT is indeed adaptable. Whenever I answer RC, it proficiently 'adapts' itself to mark my 'right' answer 'wrong'.

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Source: Beat The GMAT — Data Sufficiency | Fri Apr 16, 2010 5:38 pm

liferocks: Legendary Member; Posts: 576; Joined: Sat Mar 13, 2010 8:31 pm; Thanked: 97 times; Followed by:1 members

by liferocks » Fri Apr 16, 2010 5:46 pm

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

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deepesh.gupta: Senior | Next Rank: 100 Posts; Posts: 43; Joined: Sun Mar 21, 2010 4:32 am; Thanked: 1 times

by deepesh.gupta » Sat Apr 17, 2010 8:38 am

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

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pradeepkaushal9518: Legendary Member; Posts: 1309; Joined: Wed Mar 17, 2010 11:41 pm; Thanked: 33 times; Followed by:5 members

by pradeepkaushal9518 » Sat Apr 17, 2010 1:56 pm

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?

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akhpad: Legendary Member; Posts: 809; Joined: Wed Mar 24, 2010 10:10 pm; Thanked: 50 times; Followed by:4 members

by akhpad » Sun Apr 18, 2010 8:39 am

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