Question 1099272
<pre>
The elements of the set {6,17,28,39,...,105} are terms of the 
arithmetic sequence with first term 6, last term 105, and common
difference 11.

So the nth term is {{{a[n]=a[1]+(n-1)d=6+(n-1)(11)=6+11n-11=11n-5}}}

The number of terms is found by setting 11n-5=105 and solving for n
                                        11n=110
                                          n=10 
So there are 10 terms.

Suppose the pth, qth, rth, and sth terms are the 4 distinct terms. 
Then their sum is given by: 

a<sub>p</sub> + a<sub>q</sub> + a<sub>r</sub> + a<sub>s</sub> =

(11p-5) + (11q-5) + (11r-5) + (11s-5) = 11p + 11q + 11r + 11s - 20 =

11(p + q + r + s)-20 = 11z-20 where z = p + q + r + s 

the sum z = p + q + r + s can take on any integer value from the 
smallest sum 1 + 2 + 3 + 4 = 10 to the largest sum 7 + 8 + 9 + 10 = 34
inclusive.  There are 34 integers from 1 to 34 inclusive.  There
are 9 integers from 1 to 9, inclusive that we do not want to count. 
So there are 34-9 or 25 integers from 10 to 34 inclusive.

Each integer from 10 to 34 inclusive gives a different sum when 
substituted in 11z-20, so the answer is that 25 integers can be 
represented as a sum of 4 distinct integers chosen from 
{6,17,28,39,...,105}.

Edwin</pre>