I came across the following question while studying logic and cannot find a solution for it anywhere. I am studying by myself and think I just don't know exactly the right terms to search for it online (I'm not sure it is called a logical equation so excuse the title of this question in case it isn't):
Given the proposition $P$, it's logical value is defined as $[P] = 0$, in case $P$ is false, and $[P] = 1$, in case $P$ is true.
Consider the following open sentences defined in the set of integers:
$ P_i(x): x \le 5$
$ P_{ii}(x): x \ge 3$
$ P_{iii}(x): $ x is odd
$ P_{iv}(x): x \ge 6$
How many solutions does the following equation have?
$ x = [P_i(x)] + 2 \cdot[P_{ii}(x)]+3\cdot[P_{iii}(x)]+4\cdot[P_{iv}(x)]$
I've made this jsfiddle and from there, I can count the number of solutions through a loop. In this case I've looped from 0 to 1000 and it yields 2 solutions. Though I can clearly reason it wouldn't be possible for a very large number to work here, since these are all sums of multiplications of 0s or 1s, I am having a hard time articulating exactly why. How would you go about finding the largest number possible, in this very specific case? So you wouldn't have to loop through values of X too far off from it?
Best Answer
First, let's observe that $[P_i(x)]+4\cdot[P_{iv}(x)]=1+3\cdot[P_{iv}(x)]$. We can check three cases:
$x<3$: Our equation becomes $x=1+3\cdot(x\%2)$ where $\%$ is the mod operator. Clearly, no even number can satisfy this equation, and neither can an odd number. So, $x\geq3$.
$3\leq x<6$ We can convert the equation to $x=3+3\cdot(x\%2)$. Again, no even or odd numbers can satisfy this equation, so we move on to case three.
$x\geq6$ Since we know any solution must satisfy this, we can convert the equation to $6+3\cdot(x\%2)$. So, $x=6$ and $x=9$ are solutions.
Hence, there are $\color{red}{2}$ solutions to this equation.