SOLUTION: Find 3 consecutive odd integers such that the second integer is 5 times the first integer, the third integer is 100 more than the first integer, and the sum of all 3 integers is 41
Algebra ->
Customizable Word Problem Solvers
-> Numbers
-> SOLUTION: Find 3 consecutive odd integers such that the second integer is 5 times the first integer, the third integer is 100 more than the first integer, and the sum of all 3 integers is 41
Log On
Question 1015553: Find 3 consecutive odd integers such that the second integer is 5 times the first integer, the third integer is 100 more than the first integer, and the sum of all 3 integers is 415.
The equation I have is
x+5x+100+x=415 and my answers are 45,225,145
I'm pretty sure that is incorrect because they aren't consecutive. what is the correct equation and answers? Found 2 solutions by macston, stanbon:Answer by macston(5194) (Show Source):
You can put this solution on YOUR website! First integer is F.
Second integer must be F+2 (consecutive odd integers) and also 5F.
So
5F=F+2
4F=2
F=1/2
1/2 is not an integer.
.
The third integer must be F+4 (to be consecutive odd) integer and also F+100.
F+4=F+100
Obviously, this can not happen.
.
There is no solution to this problem as given.
You can put this solution on YOUR website! Find 3 consecutive odd integers such that the second integer is 5 times the first integer, the third integer is 100 more than the first integer, and the sum of all 3 integers is 415.
-----
1st:: 2x-1
2nd:: 2x+1
3rd:: 2x+3
----------
Note:: The conditions for your equations are contradictory.
They do not result in a consistent value for "x" and therefore
for the 3 consecutive odd integers.
=======
Cheers,
Stan H.
-------------
(