Question 1140609
a list of whole numbers from 1 to 50 is written on a piece of paper. all the 
multiples of 5 are then struck off from the list. what is the last digit of the
product of the remaining numbers.<pre>

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 a<sub>1</sub> = 1, n=50, and a<sub>n</sub> = a<sub>50</sub> = 50 

{{{S[n]=expr(n/2)(a[1]+a[n])}}} becomes
{{{S[50]=expr(50/2)(1+50)}}}
{{{S[50]=25(51)}}}
{{{S[50]=1275}}}

Now from that we must subtract this sum: 5+10+15+···+40+45+50, which forms 
an arithmetic sequence where a<sub>1</sub> = 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 a<sub>n</sub> = a<sub>10</sub> = 50,  

{{{S[n]=expr(n/2)(a[1]+a[n])}}} becomes
{{{S[10]=expr(10/2)(5+50)}}}
{{{S[10]=5(55)}}}
{{{S[10]=275}}}

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! <font face="wingdings">J</font>

<font size=1>I confess! I did it the hard way first, and then the easy way dawned on me and I felt stupid!</font>

Edwin</pre>