How many triangles can be constructed so that lengths of the sides are three consecutive odd integers and the perimeter is less than 1000?

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Raghunathan Narasimhan

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Maths and Physics for IIT-JEE & Physics for NEET

let the sides are x,x+2,x+4 ,x being an odd integer then , x+(x+2)+(x+4)<1000 , i.e, 3x<994 ,x being an odd integer ,max value of x can be 331.....now x can be any odd no from 1 to 331...like 1,3,5,7,..............,329,331...and for every such x,you will get 1 triangle..(options form an arithmetic progression)......331=1+(n-1)2,...
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let the sides are x,x+2,x+4 ,x being an odd integer then , x+(x+2)+(x+4)<1000 , i.e, 3x<994 ,x being an odd integer ,max value of x can be 331.....now x can be any odd no from 1 to 331...like 1,3,5,7,..............,329,331...and for every such x,you will get 1 triangle..(options form an arithmetic progression)......331=1+(n-1)2, which gives n=116...so 116 possible triangles satisfy your conditions!!!! read less
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Let the sides be x,x+2,x+4 ,x being an odd integer then , x+(x+2)+(x+4)<1000 , i.e, 3*x<994 ,x being an odd integer. Max value of x can 331 Now x can be any odd no from 1 to 329...like 1,3,5,7,..............,331 and for every such x, you will get one triangle.. From arithmetic progression......331=1+(n-1)*2,...
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Let the sides be x,x+2,x+4 ,x being an odd integer then , x+(x+2)+(x+4)<1000 , i.e, 3*x<994 ,x being an odd integer. Max value of x can 331 Now x can be any odd no from 1 to 329...like 1,3,5,7,..............,331 and for every such x, you will get one triangle.. From arithmetic progression......331=1+(n-1)*2, i.e. (n-1)*2 = 330 (n-1) = 165 n=166 which gives n=166 So 166 possible triangles read less
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Physics & Maths

165 @Neel, @Vikas, @ Raghunathan... U need to remove one case 1,3,5 as it cannot form sides of a triangle (since sum of two sides cannot be less than third side)
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Math Educator for Std.11th ,12th , Engineering Entrance and Degree Level with 11+ Years Experience

NO. OF POSSIBLE TRIANGLES = 166
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Maths and Physics for IIT-JEE & Physics for NEET

let the sides are x,x+2,x+4 ,x being an odd integer then , x+(x+2)+(x+4)<1000 , i.e, 3x<994 ,x being an odd integer ,max value of x can be 331.....now x can be any odd no from 1 to 331...like 1,3,5,7,..............,329,331...and for every such x,you will get 1 triangle..(options form an arithmetic progression)......331=1+(n-1)2,...
read more
let the sides are x,x+2,x+4 ,x being an odd integer then , x+(x+2)+(x+4)<1000 , i.e, 3x<994 ,x being an odd integer ,max value of x can be 331.....now x can be any odd no from 1 to 331...like 1,3,5,7,..............,329,331...and for every such x,you will get 1 triangle..(options form an arithmetic progression)......331=1+(n-1)2, which gives n=166...so 166 possible triangles satisfy your conditions!!!! read less
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