Question

Find all sets of three consecutive even integers whose sum is greater than 102 and less than 116

Answer 1

so three consecutive integers 2k,2(k+1),2(k+2)
there sum will be
2k+2(k+1)+2(k+2)
2k+2k+2+2k+4
6k+6
so setting the inequality
102<6k+6<116
subtracting 6 from all sides
96<6k<110
16<k<18.3333
since k is an integer
the only integers between 16 and 18.3
are 17 and 18
so
when k=17
2(17),2(17+1),2(17+2)
34,36,38
when k=18
2(18),2(18+1),2(18+2)
36,38,40
so to check are soultion we substitute
34+36+38=108 which is between 102 and 116
36+38+40=114 which is between 102 and 116
so our soultion is correct
the final answer will be
{{36,38,40},{38,40,42}}