José tore out some consecutive leaves from a book with pages numbered with consecutive integers and written on both sides of each leaf. The sum of the numbers on the torn pages is 344.

a) Determine the prime factorization of the number 344.

b) Find the sum of the first and the last number among those written on the torn pages.

c) How many pages were torn out?
Solution

a) Since 44 is a multiple of 4, then $344=4 \cdot 86=2^{3} \cdot 43$.

b) If $y$ is the number of leaves torn out, and $2 y$ is the number of pages, and $x+1$ is the first number that appears on them, we have

$$
\begin{aligned}
(x+1)+(x+2)+(x+3)+\ldots+(x+2 y) & =344 \\
2 x y+(1+2+\ldots+2 y) & =2^{3} \cdot 43
\end{aligned}
$$

Recalling that

it follows that

$$
1+2+3+\ldots+n=\frac{n(n+1)}{2}
$$

$$
\begin{aligned}
2^{3} \cdot 43 . & =2 x y+\frac{2 y(2 y+1)}{2} \\
& =y(2 x+2 y+1)
\end{aligned}
$$

Since $2 x+2 y+1$, which is the sum of the numbers on the first and last torn pages, is an odd divisor of $2^{3} \cdot 43$ greater than 1, it follows that $2 x+2 y+1=43$.
c) Since $2 x+2 y+1=43$, from the equation in the previous item, we arrive at $y=2^{3}=8$. Therefore, the number of pages torn out is $2 y=16$.