2007 SMT Algebra Round Problem 4

Problem

How many positive integers $n$ , with $n\leq2007$ , yield a solution for $x$ (where $x$ is real) in the equation $\lfloor x\rfloor+\lfloor 2x\rfloor+\lfloor 3x\rfloor=n$ ?

Solution

Let's try to find solutions for $n$ If $n=1$ , then we have a solution when $x=0.35$ . If $n=2$ , then we have a solution when $x=0.5$ . If $n=3$ , then we have a solution $x=0.9$ . If $n=6$ , then we have a solution $x=1$ . If $x<1$ , we have $n\leq3$ and if $x\geq1$ , then we have $n\geq6$ , so $n=4$ or $5$ do not have a solution. If $n\geq6$ , then just add $\left\lfloor \frac{n}{6}\right\rfloor$ to the solution to the remainder of $\frac{n}{6}$ and if the remainder of $\frac{n}{6}$ is $4$ or $5$ , then there is no $x$ that satisfies the equation. So, we have $4$ possibilities out of every $6$ numbers, so we can get the number of solutions by doing $4\times\left\lfloor \frac{2007}{6}\right\rfloor=1336$ and then getting that the remainder of $\frac{2007}{6}$ is 3 and so we have $n$ 's where when divided by 6 gives remainder of $1,2,$ and $3$ , so they all have a solution of $x$ , and so we add $3$ to $1336$ to get $\boxed{\mathrm{1339}}$ .

~Yuhao2012

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2007 SMT Algebra Round Problem 4

Problem

Solution