Question 1099272
<br>Tutor Edwin has, as usual, provided an excellent solution to your problem.<br>
I would approach the problem a bit differently; so let me show you how I would solve the problem.<br>
You might find one solution method or the other more to your liking; I'm not saying one method is better than the other....<br>
The numbers are
6, 17, 28, ..., 83, 94, 105<br>
They form an arithmetic sequence with common difference 11.<br>
The smallest sum possible from 4 of these number is {{{6+17+28+39 = 90}}};
the largest sum possible is {{{72+83+94+105 = 354}}}.<br>
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.<br>
That is, the sums that can be formed are the numbers in the arithmetic sequence
90, 101, 112, ..., 332, 343, 354<br>
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:
{{{354-90 = 264}}}
{{{264/11 = 24}}}
{{{24+1 = 25}}}