For a positive real number $ [x] $ be its integer part. For example, $[2.711] = 2, [7] = 7, [6.9] = 6$. $z$ is the maximum real number such that [$\frac{5}{z}$] + [$\frac{6}{z}$] = 7. Find the value of$ 20z$.
Problem
Source: BdMO 2020 Final Primary Category Problem 5
Tags: floor function, algebra
30.08.2020 23:23
Please define "[n]"? It is not the same has floor function. It is the fractional part?
30.08.2020 23:46
Thank you,it is updated
31.08.2020 00:18
add the space between "of" and "$20z$"... sorry for being so critical
31.08.2020 02:50
$z=1.5$ so the answer is $30$. This is the largest since anything bigger would make $[6/z]<4$ which is absurd.
10.10.2020 13:03
A bit too easy for an olympiad... We see that this function is just the floor function... Since $\lfloor{\frac{5}{z}}\rfloor$ and $\lfloor{\frac{6}{z}}\rfloor$ must be integers, we see that: $$\lfloor{\frac{5}{z}}\rfloor = 3$$$$\lfloor{\frac{6}{z}}\rfloor = 4$$ This implies that $z$ could be ranging from $\frac{5}{4}$ to $\frac{6}{4}$ (inclusive). Since we want the maximum, $z = \frac{6}{4}$, which implies that $20z = \boxed{30}$.