Question 1196182: Find three consecutive odd integers such that the sum of two times the first, three times the second, and one times the third is 100. List the numbers in order from smallest to largest.
Found 3 solutions by Alan3354, greenestamps, josgarithmetic:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Find three consecutive odd integers such that the sum of two times the first, three times the second, and one times the third is 100. List the numbers in order from smallest to largest.
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2n + 3(n+2) + (n+4) = 100
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
While it is likely that you are supposed to use formal algebra to solve this problem, you can get some good mental exercise, and good practice with problem solving, by solving this problem using logical analysis and a bit of approximation.
In any string of three consecutive odd integers, the middle one is the average of all of them.
So 2 times the first, plus 3 times the second, plus 1 times the third, is approximately equal to 6 times the second.
Since that sum is equal to 100, we have that (approximately) 6 times the middle number is 100.
100/6 = 16 2/3; then, since the numbers are odd integers, the middle number should be 17. So the numbers are 15, 17, and 19.
ANSWER: 15, 17, and 19
CHECK: 2(15)+3(17)+19 = 30+51+19 = 100
Answer by josgarithmetic(39620)  (Show Source):
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