Bdmo Junior 2022 problem 5

Let \lfloor x \rfloor denotes the smallest intezer \le x. Example :- \lfloor 2.1 \rfloor = 2 , \lfloor -2.3 \rfloor = -3 , \lfloor 6 \rfloor = 6

Solve the equation for all real x ,
\lfloor x \rfloor ^ 3 - 7 \lfloor x +\frac{1}{3} \rfloor = -13

[BDMO National Junior 2022 P5]

Note that \lfloor{x+\frac{1}{3}}\rfloor = \lfloor{x}\rfloor or \lfloor{x}\rfloor +1

Now, you can consider each case and do algebra. Also remember that,
-13 = 1*-13 or -1*13 … as i can remember, when you do the second case, you get something equal to -6. I bashed and consider each case like -6= 1*-6,2*-3,3*-2,6*-1

2 Likes

Then the case becomes like y(y^2-7)=-13 where y = \lfloor x \rfloor

Case 1 : |y| = 1
Then y^2 - 7 = -6 so y = 1

Case 2: |y| = 2
Then y^2-7 = -3 so y = 2

Case 3 : |y| = 3
Then y^2-7 = 2 so y=-3

Case 4: |y| = 6
Then y^2-7 = 31 so no y

Possible solution for y = {1,2,-3}
Now we have to find x