### Instruction: A math textbook with a double-digit number of pages is split into sections. Each section is exactly $12$ pages long, with the exception of the epilogue, which is $11$ pages long. Every page belongs to a section. Furthermore, on the bottom of each $5$th page, a trivia fact is presented on the bottom of the page, starting from the fifth page. If a trivia fact appears on the bottom of the second-to-last page, then how many pages does the textbook have?### Response: I want to find the total number of pages in the textbook, which I'll call $p$.
I also want to use the fact that a trivia fact appears on the bottom of the second-to-last page, which means that the second-to-last page is a multiple of $5$.
Let me write that as an equation: $p - 2 = 5k$, where $k$ is some positive integer.
I can rearrange this equation to get $p = 5k + 2$.
Now I have a formula for the total number of pages, but I need to make sure it satisfies the other conditions of the problem.
One condition is that each section is exactly $12$ pages long, except for the epilogue, which is $11$ pages long.
This means that the total number of pages must be $11$ more than a multiple of $12$, because the epilogue is the extra $11$ pages at the end.
Let me write that as another equation: $p = 12m + 11$, where $m$ is some positive integer.
Now I have two equations for $p$, and I need to find a common solution that makes both equations true.
I can try to solve for $m$ in terms of $k$ by equating the two expressions for $p$: $5k + 2 = 12m + 11$.
Subtracting $2$ from both sides gives me $5k = 12m + 9$.
Dividing both sides by $5$ gives me $k = \frac{12m + 9}{5}$.
Since $k$ has to be a positive integer, this means that the numerator has to be divisible by $5$.
I can try different values of $m$ until I find one that makes the numerator divisible by $5$.
If I start with $m = 1$, I get $k = \frac{21}{5}$, which is not an integer.
If I try $m = 2$, I get $k = \frac{33}{5}$, which is still not an integer.
If I try $m = 3$, I get $k = \frac{45}{5} = 9$, which is an integer.
So $m = 3$ and $k = 9$ are possible values that make both equations true.
To find the total number of pages, I can plug in either $m = 3$ or $k = 9$ into either equation.
For example, using $k = 9$, I get $p = 5k + 2 = 5 \times 9 + 2 = 47$.
Alternatively, using $m = 3$, I get $p = 12m + 11 = 12 \times 3 + 11 = 47$.
Either way, I get the same answer, which is $p = 47$.