Question 378034: Find all sets of three consecutive odd integers whose sum is at least 47 less than 4 times the third integer.
Please help! Thank you.
Found 2 solutions by robertb, mananth:
Answer by robertb(5830)   (Show Source):
You can put this solution on YOUR website!  ,
 ,
 ,
 .
 . Then the possible sets are
{37, 39, 41},
{35, 37, 39},
(33, 35, 37},
(31, 33, 35},...
............
{3,5,7},
{1,3,5},
{-1, 1, 3},
{-3, -1, 1},
{-5, -3, -1},....(There are infinitely many!)
Answer by mananth(16946)  (Show Source):
You can put this solution on YOUR website! let x, x+2, x+4 be the integers.
their sum = x+x+2+x+4
=3x+6
...
47 less than 4 times the third integer = 4(x+4)-47
...
3x+6 <= 4(x+4)-47
3x+6<=4x+16-47
3x+6<=4x-31
-6
3x+6-6 <=4x-31-6
3x<=4x-37
-4x
3x-4x<=4x-4x-37
-x<=-37
/-1
x>=37
37,39,41,is the first set
...
m.ananth@hotmail.ca
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