If N is a positive integer, what is the last digit of 1! +

Generally the last digit of \(1! + 2! + ... +N!\) can take ONLY 3 values:
A. N=1 --> last digit 1;
B. N=3 --> last digit 9;
C. N=any other value --> last digit 3 (N=2 --> 1!+2!=3 and for N=4 --> 1!+2!+3!+4!=33, \(N\geq{4}\) the terms after N=4 will end by 0 thus not affect last digit and it'll remain 3).

So basically question asks whether we can determine which of three cases we have.
(1) N is divisible by 4 --> N is not 1 or 3, thus third case. Sufficient.

(2) (N^2 + 1)/5 is an odd integer --> N is not 1 or 3, thus third case. Sufficient.

Answer: D.

Hope it helps.
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Even if the generalization doesn't occur to you, take some values to see if there is a pattern:
1! = 1
1! + 2! = 3
1! + 2! + 3! = 9
1! + 2! + 3! + 4! = ...3
1! + 2! + 3! + 4! + 5! = ....3 (because 5! is 120 so it doesn't contribute in last digit)
As we go higher, 6!, 7! etc, each will end with 0 because it will have a 5 and a 2.
So we only have to rule out N=1 and N=3.

statement 1: n can = 4, 8, 12, etc

4! + 3! + 2! + 1!

24 + 6 + 2 + 1 = 33 last digit is 3
If x! and x>=5 then the number will always end in zero

Essentially you will always be adding 4! + 3! + 2! + 1! and always have a value of 3 as the last digit

Nice Bunuel.
Figuring out the generalization about the last digit of 1!+2!+...+N! is the tricky part.

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

(C) 2008 GMAT Club - m15#15

* Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
* Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
* BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
* EACH statement ALONE is sufficient
* Statements (1) and (2) TOGETHER are NOT sufficient

\(1! + 2! + ... + N!\) ends with 3 if \(N \gt 3\) . Starting from 5! each term in the sum ends with 0, so if \(N \gt 3\) , the sum looks like \(1! + 2! + 3! + 4! + \text{(something ending with 0) } = 33 + \text{ (something ending with 0) } = \text{ integer ending with } 3\) .

Statement (1) by itself is sufficient. From S1 it follows that \(N\) is 4 or bigger.

Statement (2) by itself is sufficient. From S2 it follows that \(N\) is not 1 or 3. \(N\) is either 2 or larger than 3. In either case, sum \(1! + ... + N!\) ends with 3.
The correct answer is D

Can some one please explain the solution for both cases?

Here, is the last digit 0 for all N > 5?

hmm! after i wrote it down on paper i understood the second part

N^2 + 1 = 5 (odd number)

n=1 does not work

n= 2, then the odd number is 1. So n can be ==2.
Then last digit is 3.

N is not equal to 3.

Here N cannot be 4. for all other numbers, the sequence ends with 3

From (A), take up to 4! (as divisible by four)

1 + 2 + 6 + 24 - Add and the last digit 3, From 5! onwards it is 60 + 120 + .. all of these have 0 at the end

so the last digit will be 3


n^2 must end in 4

(2^2 + 1)/5 = 1 But 1! + 2! has last digit as 3

(8^2 + 1)/5 = 13 But 1! + ... + 8! has last digit as 3, so enough.

Answer is D.
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I think the answer would be A

Lets analyse statement 1 first.

1 != 1*1
2 != 2*1
3 !=3*2*1

3x2x1+2x1+1*1/4= 2
Hence 3!
Hence Sufficient

Lets analyse statement 2 now

N/2+1/5 is a odd number
5(N+2)/2
so you can subsititude N=4, which will get you 15, its odd, so sufficient
however if you subsititude N=5 which will you get you 45/2 non intenger so insufficient

Guys, do you assume that there are no numbers between 2! and N!?
I also don't understand this point: N=1 --> last digit 1. Can someone explain it to me? Please.

I guess we assume that there are (N - 2) numbers between 2 and N. We have to find the sum of the factorials of each of these (N - 2) numbers.

But if we want to find the last digit of 1! (that is what Bunuel meant by N = 1) then the units digit will be 1.
If we are asked to find the sum of factorials of the first three numbers (N = 3 in this case) then the units digit will be 9.
And for all other numbers (any other value of N), the sum of the factorials of the numbers till N will have 3 as the units digit.

Nice explanation bunuel

Could you please explain the bold face part? how is having 0 not affecting last digit and still remain 3? Wouldn't the last digit be 0?

Just plug numbers to check.

If N=4 then \(1! + 2! + 3!+4!=1+2+6+24=33\);
If N=5 then \(1! + 2! + 3!+4!+5!=(1+2+6+24)+120=33+120=153\);
If N=6 then \(1! + 2! + 3!+4!+5!+6!=(1+2+6+24)+120+720=33+120=873\);
...
As you can see the units digit if N=4 is 3, now the terms after N=4 (5!, 6!, ...) will have the units digit of 0 (since they all have the product of 2 and 5 in them), so adding those terms won't affect the units digit.

Hope it's clear.
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BB i am confused here...

A: N = 4 , u said the 4! = 4*3*2*1 = 24, 24+6+2+1 = 33
N= 8, tthe value will be someting else, but the last digit is 0
so the last digit is always 3....

B: N = 2, then the integer from B is 5/5 = 1, then unit digit is ( 2!+1!) 3,
N = 3, does not get odd integer
N = 4, does not get integer
N = 5, does not get integer
N = 6, does not get integer
N = 7, does not get odd integer
N = 8, does not get integer....
here, since N>4, last digit is 0, the value is 3...

very tricky.. i gave it for B, but the anser is D

First of all I'm not BB. BB is completely different person: members/member-3.html

Next answer to this question is D, not B.

Again, the units digit of \(1! + 2! + ... +N!\) can take ONLY 3 values 1, 9, or 3:
A. If N=1 then the units digit is 1;
B. If N=3 then the units digit is 9;
C. If N=any other value then the units digit is 3.

So basically question asks whether we can determine which of three cases we have.

(1) N is divisible by 4 --> N is not 1 or 3,because neither 1 nor 3 is divisible by 4, hence we have case C. For case C the units digit of \(1! + 2! + ... +N!\) is 3. Sufficient.

(2) (N^2 + 1)/5 is an odd integer --> again N is not 1 or 3, because neither 1 nor 3 gives odd value for (N^2 + 1)/5, hence we have case C. For case C the units digit of \(1! + 2! + ... +N!\) is 3. Sufficient.

Answer: D.

In either of statement it doesn't matter what the actual value of N is. Since we can get that it's not 1 or 3 then in all other cases the units digit of \(1! + 2! + ... +N!\) will be 3.
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Perfectly clear and i can't thank you enough Bunuel!!!!

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