If n is a positive integer, what is the units digit of the sum of the following series:1! + 2! + ... + n!? (The series includes every integer between 1 and n, inclusive) (1) n is divisible by 4. (2) n² + 1 is an odd integer.
A. Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not. B. Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not. C. Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though NEITHER statement BY ITSELF is sufficient. D. Either statement BY ITSELF is sufficient to answer the question. E. Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, meaning that further information would be needed to answer the question.
(D) In this question, we need to look for a pattern in the units digit of the sum of factorials for positive integers. Start with 1! + 2! = 3. Then since 3! = 6 add this to the previous sum and get 9. 4! = 24. Add this to the previous sum: 33. 5! = 120. Adding this: 153. 6! = 720…
Let’s stop here, because the pattern should now be clear. Every additional number that will be added to the series will have a units digit of zero (because it has 2 and 5 as factors and therefore 10).
So for all n > 3, the units digit of the sum will be 3. Now, let's look at the statements.
Statement (1) tells us that n is a multiple of 4. Since n must be greater than 3, we know the units digit of the sum of the series will always be 3, so Statement (1) is sufficient.
From Statement (2), we can determine that n cannot be either 1 or 3. It is possible that n = 2, in which case: 1! + 2! = 3. Any other number that satisfies the statement is greater than 4, in which case the sum 1! + ... + n! will have a units digit of 3 as well. So Statement (2) is also sufficient because the units digit is always 3.
Since both statements are sufficient individually, the correct answer is choice (D). 
The explanation says that 1! + 2! = 3. But I thought 1! + 2! = 1 + 3 = 4
Thank you.
