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Question 76533: Find all sets of three consecutive multiples of 11 for which the sum of the lesser numbers is greater then 100 and the sum of the two greater numbers is less than 200
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Find all sets of three consecutive multiples of 11 for which the sum of the lesser numbers is greater then 100 and the sum of the two greater numbers is less than 200
:
Let x = the multiplier
The numbers 11x, (11x+11), (11x+22)
Find x;
"the sum of the two greater numbers is less than 200"
(11x+11) + (11x+22) < 200
22x + 33 < 200
22x < 200 - 33
22x < 167
x < 167/22
x < 7.5
x = 7 (has to be an integer)
:
Gives us the numbers 77, 88, 99 max
:
Sum of the two smallest: 165
Sum of the two larger: 187
:
Find x using:
11x + (11x+11) > 100
22x + 11 > 100
22x > 100 - 11
22x > 89
x > 89/22
x > 4.04
x = 5
:
Gives us 55, 66, 77 min
:
3 sets would be 55,66,77; 66,77,88; 77,88,99
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