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\( \lfloor t\rfloor = 3t + 4 - \lfloor 2t \rfloor\)

Looking at this equation, what can we conclude about t? \(\lfloor t \rfloor, 4, \lfloor 2t \rfloor\), are all integers, because that is the property of floors.

Here is a little trick, now we know 3t is an integer, so t = some integer / 3.

\(\because\)  \(t\equiv 0 \pmod 3\), or \(t \equiv 1 \pmod3\) or \(t \equiv 2 \pmod3\)

\(\therefore\) t either has a decimal part of \(0, \frac{1}{3}, \frac{2}{3}\).

Lets split this in three cases:

Case 1:

t has a decimal part of 0 or  \(t=\lfloor t \rfloor\).

This case is realtively simple, replacing all floors with t gives us,

\(t = 3t+4-2t\), or \(t=t+4\). We see that this is impossible because \(0\neq4\).

Case 2:

t has a decimart part of 1/3.

We see that \(t=\lfloor t\rfloor+\frac{1}{3}\). or \(\lfloor t\rfloor=t-\frac{1}{3}\)

Since the decimal part of t is 1/3, then the decimal part of 2t doesn't exceed 1, so\(\lfloor 2t\rfloor = 2t-\frac{2}{3}\)

Substituting, \(\lfloor t\rfloor=3t+4-2t+\frac{2}{3}\). Subbing our conversion in, and simplifying, \(t-\frac{1}{3}=t+\frac{14}{3}\)\(0\neq5\), we see that this case has no solution too.

Case 3:

t has a decimal part of 2/3.w

\(t=\lfloor t\rfloor+\frac{2}{3}\), or \(\lfloor t\rfloor=t-\frac{2}{3}\).

We need to be careful here, because \(2t=\lfloor 2t\rfloor+\frac{4}{3}\), and 4/3 exceeds the next number so \(\lfloor 2t\rfloor=2t-\frac{1}{3}\).


\(t-\frac{2}{3}=3t+4-2t+\frac{1}{3}\)\(t-\frac{2}{3}=t+\frac{13}{3}\)\(0\neq5\). We once again see there is no solution.

This means there is no solution, to \( \lfloor t\rfloor = 3t + 4 - \lfloor 2t \rfloor\).

P.S, This can also be seen through a graph, where \(y = \lfloor x\rfloor + \lfloor 2x \rfloor\), and \(y=3x+4\)