There is an easy way and a hard way:

Here's the easy way:

The last digits of the whole numbers from 1 to 50 (inclusive) make up 5 groups
of 1,2,3,4,5,6,7,8,9,0.  When we take away the multiples of 5, we take away
those ending in 0 or 5, and that leaves 5 groups of 1+2+3+4+6+7+8+9 = 40, and
adding any whole number of groups of 40 will end up having a 0 for the last
digit of the sum.

Answer: 0
  
Now for the hard way:

First we find the sum of 1+2+3+···+48+49+50

The formula for the sum of the first N whole numbers, since they form an
arithmetic sequence where a1 = 1, n=50, and an = a50 = 50 

 becomes




Now from that we must subtract this sum: 5+10+15+···+40+45+50, which forms 
an arithmetic sequence where a1 = 5. But how am I going to find out
what to substitute for n? This way:

If we divide every number in this series

5+10+15+···+40+45+50

by 5, we get this series:

1+2+3,···,8+9+10

and since we know that has 10 terms, so does the first one!  So n=10,
and an = a10 = 50,  

 becomes




So to get the final answer the hard way, we subtract:

1275-275 = 1000, which ends in 0.   <-- final answer

The easy way is better! J

I confess! I did it the hard way first, and then the easy way dawned on me and I felt stupid!

Edwin

First 10 numbers of the sequence are 


    1  2  3  4  5  6  7  8  9  10


When you take off 5 and 10 from this list, you will get the sequence


    1  2  3  4     6  7  8  9


The product of these numbers is  1*2*3*4*6*7*8*9 = 72576.


The last digit of this product is 6.


It gives you an idea that the last digit of the number under the question is   = 6.     ANSWER