Question 1099272: How many different integers can be represented as a sum of four distinct numbers chosen from the set ?
Found 2 solutions by Edwin McCravy, greenestamps:
Answer by Edwin McCravy(20056)   (Show Source):
You can put this solution on YOUR website!
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 
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:
ap + aq + ar + as =
(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
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
Tutor Edwin has, as usual, provided an excellent solution to your problem.
I would approach the problem a bit differently; so let me show you how I would solve the problem.
You might find one solution method or the other more to your liking; I'm not saying one method is better than the other....
The numbers are
6, 17, 28, ..., 83, 94, 105
They form an arithmetic sequence with common difference 11.
The smallest sum possible from 4 of these number is  ;
the largest sum possible is  .
Because the difference between terms of the sequence is a constant 11, the only sums possible using 4 of the numbers are the numbers that are some multiple of 11 larger than 90 and less than or equal to 354.
That is, the sums that can be formed are the numbers in the arithmetic sequence
90, 101, 112, ..., 332, 343, 354
There is a simple standard calculation for finding the number of numbers in that sequence -- (last term minus first term); divided by the common difference; plus 1:
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