Data Sufficiency

Hi vipulgoyal,

What is the source of this question? I ask because I have the same issue with it that you do....

Your immediate question is why N cannot be 2. Based on the design of the prompt, it 'appears' that we're meant to take the sum of....

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

Based on this "phrase", the implication is that N must be greater than 2. If this were an Official GMAT question, there would be some wording that stated "N is a positive integer greater than 2" so that there would be no confusion or ambiguity.

GMAT assassins aren't born, they're made,
Rich

vipulgoyal 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.

Target question: What is the last digit of 1! + 2! + ... +N!?

Given: N is a positive integer

Statement 1: N is divisible by 4
So, N can equal 4, 8, 12, 16, 20, etc.

Let's test some values of N

N = 1: 1! = 1
N = 2: 1! + 2! = 1 + 2 = 3
N = 3: 1! + 2! +3! = 1 + 2 + 6 = 9
N = 4: 1! + 2! +3! + 4! = 1 + 2 + 6 + 24 = 33
N = 5: 1! + 2! +3! + 4! + 5! = 1 + 2 + 6 + 24 + 120 = 153
N = 6: 1! + 2! +3! + 4! + 5! + 6! = 1 + 2 + 6 + 24 + 120 + 720= 873
N = 7: 1! + 2! +3! + 4! + 5! + 6! + 7! = 1 + 2 + 6 + 24 + 120 + 720 + 5040 = 5913
.
.
.
As we can see, for every value of N greater than 3, the units digit is always 3
Since N > 4, we can be certain the sum will have 3 as its units digit.
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Aside: For more on this idea of plugging in values when a statement doesn't feel sufficient, you can read my article: https://www.gmatprepnow.com/articles/dat ... lug-values


Statement 2: (N^2 + 1)/5 is an odd integer
There are several values of N.
N could equal 2, or 8 or 12 or 18 or 22 or...
For all of these possible values of N, we can be certain the sum will have 3 as its units digit.
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer = D

Cheers,
Brent

vipulgoyal wrote:Hi Brent, If n = 2 , how could unit digit be 3, 1! +2! +2! = 5

If N = 2, we get: 1! + 2! = 1 + 2 = 3

In your calculation, you have an extra 2!

Cheers,
Brent

vipulgoyal wrote:Hi Brent, If n = 2 , how could unit digit be 3, 1! +2! +2! = 5

Hi vipulgoyal,

You may have forgotten that, in your original post, you (correctly) wrote the following:

C. N=any other value --> last digit 3 (N=2 --> 1!+2!=3....)

Cheers,
Brent

vipulgoyal wrote:what is the last digit of 1! + 2! + ... +N!?

dont know seems this discussion going abase , i believe one 2! is already there and when we put n = 2 , second 2! comes there

Having said that"C. N=any other value --> last digit 3 (N=2 --> 1!+2!=3....)" is OE by expert and exactly in line with my query , Stem doesn't says that N must have another(other then 2) value, I still believe this Q is flawed ??

I wouldn't say the question is flawed.

The idea with the formula (1! + 2! + ... +N!) is that we keep adding successive factorials up to N!
So, if N = 1, we get 1!
If N = 2, we get 1! + 2!
If N = 3, we get 1! + 2! +3!
We don't assume that the 1! and 2! are in the sum regardless of the value of N.

This is no different that the explanation for Problem Solving question #172 in the Official Guide.
It explains how to find the sum of the first n positive integers.
It says: 1 + 2 + ... + n = (n)(n+1)/2
So, to find the sum of the first 2 positive integers, we plug in n = 2 to get:
Sum = (2)(2+1)/2 = 3
Notice that, when n = 2, we aren't talking about the sum 1 + 2 + 2 (even though the formula may suggest this)

This is analogous to the question you are asking.

Cheers,
Brent