Question 1169664
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&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;From the first glance, the problem seems to be hard, &nbsp;but . . . actually . . . <U>it is a joke problem</U>.



<pre>
As soon as the last digit of n! is zero, it does not influence more on  the last digit of the sum . . . 


So, the last requested digit comes from

    1! + 2! + 3! + 4! + 5! = 1 + 2 + 6 + 24 + 120 


and is,  <U>T H E R E F O R E</U>,   3.


<U>ANSWER</U>.  The last digit of  1! + 2! + 3! + . . .  + 2006!   is   {{{highlight(3)}}}.
</pre>

Solved.