SOLUTION: Four time the smallest of three consecutive odd integers is 236 more than the sum of the other two intergers. Find the integers.
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Question 271334: Four time the smallest of three consecutive odd integers is 236 more than the sum of the other two intergers. Find the integers. Answer by checkley77(12844) (Show Source):
You can put this solution on YOUR website! Let x, x+2 & x+4 be the 3 odd integers.
4x=236+(x+2)+(x+4)
4x=236+2x+6
4x-2x=242
2x=242
x=242/2
x=121 ans. for the smallest integer.
121+2=123 for the middle integer.
121+4=125 for the largest integer.
Proof;
4*121=236+123+125
484=484
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