What is 1! + 2! + ... + 10! ?

You can apply the formula above: \(Sum=\frac{first+last}{2}*number \ of \ terms\), the mean multiplied by the number of terms, for an evenly spaced set (aka arithmetic progression). But as in the sequence given (1!, 2!, 3!, ..., 10!), the difference between any two successive terms is not the same, so we don't have an evenly spaced set (arithmetic progression) and thus cannot apply this formula.

Check this thread for more: https://gmatclub.com/forum/math-number- ... 88376.html and also this thread: https://gmatclub.com/forum/sequences-pr ... 01891.html

As for the solution:

What is the value of \(1! + 2! + ... + 10!\)?

A. 4,037,910
B. 4,037,913
C. 4,037,915
D. 4,037,916
E. 4,037,918


For any integer \(n\) greater than or equal to 5, the units digit of \(n!\) is zero. This is because, for \(n \geq 5\), \(n!\) will contain at least one 2 and at least one 5 as factors. When multiplied together, these produce a trailing zero. Hence, the terms from \(5!\) through \(10!\) each have zero as their units digit. Adding up the remaining terms, we'd have \(1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33\). Therefore, we see that the entire sum will have a units digit of 3 + 0 = 3. Only option B provides a number with 3 as its units digit.

Alternatively, we can do the following:

\(1! + 2! + ... + 10! = 1 + (2! + ... + 10!)\) = odd + even = odd.

\(1! + 2! + ... + 10! = 3 + (3! + ... + 10!)\) = integer divisible by 3.

Among the listed choices we are looking for odd integers divisible by 3. Only choice B fits.


Answer: B