Question 420018
Let a, b, and c be the lengths of the sides of the triangle. So the perimeter is a+b+c. Because we know that the perimeter is between 10 and 24 (inclusive), we know that 


{{{10<=a+b+c<=24}}}



Moreover, we know that the three sides are consecutive integers. So a=x, b=x+1, and c=x+2 for some integer x. So replace a, b, and c to get



{{{10<=x+(x+1)+(x+2)<=24}}}



{{{10<=3x+5<=24}}} Combine like terms



{{{10-5<=3x<=24-5}}} Subtract 5 from all sides



{{{5<=3x<=21}}} Combine like terms.



{{{5/3<=x<=21/3}}} Divide all sides by 3 to isolate x



{{{2<x<=21/3}}} Divide 5 by 3 to get 2.6667 and round down to get 2. Since 2.667 is larger than 2 and x > 2.667, this means that x > 2. 



{{{2<x<=7}}} Divide 21 by 3 to get 7



So the possible values for a range from 2 (non inclusive) to 7 (inclusive). So the values of a are: 3, 4, 5, 6, 7



Note: the values of b and c depend on a.




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