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

Question

Jeffrey Gates

Mathematics

8 November, 12:19

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

+5

Answers (1)

Know the Answer?

Perla Quinn · 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
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}}