There were 297 numbers on the page number of a novel. How many pages were there in this novel?

Answer 1

This problem could be solved through mathematical methods. Assuming that the novel has $n$pages, then each page has $p$numbers, where $1'le p 'le n$.

According to the page number of the question, a total of $297$numbers can be listed as follows:

$$n\times p + n - 1 = 297$$

To simplify it:

$$n(p+1) = 297 - 1 = 296$$

Since $n$is an integral,$p+1$must be a multiple of $296$. At the same time, since $1'le p 'le n$,$p+1$must be a multiple of $12' ldotsn'$.

Therefore, the following restrictions can be obtained:

$$p+1> text {is a multiple of $1$but not a multiple of $2$} p+1> text {is a multiple of $2$but not a multiple of $3$}& ldots p+1> text {is a multiple of $n$but not a multiple of $n-1 $}$$

According to these constraints, the value range of $p+1$can be obtained:

$$135791113\ldotsn$$

Substituting these values into the equation $n(p+1) = 297 - 1$gives:

$$n(n+1) = 297 \times (n+1)$$

To simplify it:

$$n^2 + n - 296 = 0$$

By solving this second order equation, one could get:

$$n = \frac{296\pm\sqrt{296^2-4\times1\times296}}{2\times1} = \frac{296\pm294}{2}$$

Since $n$is an integral number,$n$can only take two values:

$$n = 44 n = 43$$

So this novel has a total of $44$or $43$pages.