11/[x] + 1/[2x] = {x} +1/3
1/3 = 1/[x] +1/[2x] - {x}
multiplying 3 on both sides
1= 3/[x] + 3/[2x] - 3{x}
left side is integer
3{x} should also be an integer
so {x} = 0 or 1/3 or 2/3
so you get three cases
is anything wrong till here?
Solve for x
1/[x] + 1/[2x] = {x} + 1/3
[x] - greatest integer function
{x} - fractional part of x
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9 Answers
62one thing is obvious the fractional part will be a rational number..
so u can start by taking {x} = p/q
where p/q < 1
1623{x} should also be an integer?
why!!
This assumption i am not sure if it is correct!!
how do u say that 3/[x] + 3/[2x] is an integer?
624/3 is the only solution..
try taking..
{x} = 1/[x] +1/[2x] - 1/3
use the inequality that it is >0
u will get it very simply :)
62ok i will post the solution i had in mind.. may be i did hadbadi me gadbadi..
then we can see what went wrong :)
62{x} = 1/[x] +1/[2x] - 1/3
{x}>=0
1/[x] +1/[2x] - 1/3>0
Case1 {x}<1/2
[2x]=2[x] = 2k let
1/[x] +1/[2x] - 1/3 = 1/k+1/2k -1/3 =3/(2k) -1/3 >0
3/(2k) >1/3
k<4.5 so k is {1,2,3,4}
Case2 {x}>=1/2
[2x]=2[x] +1 = 2k+1 let
1/[x] +1/[2x] - 1/3 = 1/k+1/(2k+1) -1/3 >0
1/k+1/(2k+1) > 1/3
k<4.5 so k is {1,2,3,4}
So we have that [x] = 1 or 2 or 3 or 4
now i guess it is simple?