Tricky modulus

There are two main ways to solve this.

SOLUTION ONE

The more orthodox way is to split into cases. Our cases are bounded by the places where the function inside an absolute value switches from positive to negative. Here, we have bounds $x\le -1$ , $-1\le x\le \frac { 3 }{ 2 }$ , $\frac { 3 }{ 2 } \le x\le 5$ , and $5\le x$ .

case 1: $x\le -1$

While $x$ satisfies this inequality, our given equation can be rewritten as $2x+3+x-1+5+x=0.99$ , or just $4x-7=0.99$ . If we approximate $0.99\approx 1$ , simple algebra yields $x\approx 2$ . However, notice that this does not satisfy our assumption that $x\le -1$ , so there are actually no solutions in this range!

case 2: $-1\le x\le \frac { 3 }{ 2 }$

We rewrite the given equation as $3-2x+x-1+5-x=0.99$ and do the same process as the previous case. We end up with $x\approx 3$ , which again violates our assumption!

case 3: $\frac { 3 }{ 2 } \le x\le 5$

We rewrite the given equation as $2x-3+x+1+5-x=0.99$ , and end up with $x\approx -1$ by our process. Yet again our value violates the assumption!

case 4: $5\le x$

We rewrite the given equation as $2x-3+x+1+x-5=0.99$ , and end up with $x\approx \frac { 7 }{ 4 }$ . In this final case, the assumption is violated, and we are left with no solutions.

Something that tripped me up was the last bit of the question, which asks for $n!$ and not $n$ itself.

$0!=\boxed{1}$

SOLUTION TWO

We can deduce logically that the minimum point of the left side of the equation is at $x=\frac { 3 }{ 2 }$ . At this point, the value of the left side of the equation is $6$ , which, being greater than $0.99$ , tells us there are no solutions. I will now add some proof to this logical conclusion:

This requires a bit of calculus knowledge. The derivative of $\left| x \right|$ is $\frac { \left| x \right| }{ x }$ . Using this in conjunction with simple chain rule, the derivative of the left side of the equation is

$\frac { 2\left| 2x-3 \right| }{ 2x-3 } +\frac { \left| x+1 \right| }{ x+1 } +\frac { -\left| 5-x \right| }{ 5-x }$ or $2sign\left( 2x-3 \right) +sign\left( x+1 \right) -sign\left( 5-x \right)$ where $sign\left( x \right)$ returns $1$ for positive $x$ , $-1$ for negative $x$ , and is undefined for $x=0$ .

In our cases from solution 1, we see cases 1 and 2 have negative slope and cases 3 and 4 have positive slope. Since the left side of the equation is continuous, the absolute minimum point will be at the point between the positive and negative slopes, at the boundary of cases 2 and 3. This $x$ -value is, not surprisingly, $\frac { 3 }{ 2 }$ . Again, $x=\frac { 3 }{ 2 }$ returns a value too great for the right side of the equation, so we have no solutions.

$0!=\boxed{1}$

I plotted a quick graph of the function in Microsoft Excel, and based on the graph, there is no value of x for which the function is 0.99. So, $n=0$ , which implies that $n!=1$ .