[Algebra]
Problem: Digital images consist of a very large number of equally spaced dots called pixels The resolution of an image is the number of pixels/cm in each of the horizontal and vertical directions.

Thus, an image with dimensions $10 \mathrm{~cm}$ by $15 \mathrm{~cm}$ and a resolution of 75 pixels/cm has a total of $(10 \times 75) \times(15 \times 75)=843750$ pixels.

If each of these dimensions was increased by $n \%$ and the resolution was decreased by $n \%$, the image would have 345600 pixels.

Determine the value of $n$.

Solution: When the dimensions were increased by $n \%$ from 10 by 15 , the new dimensions were $10\left(1+\frac{n}{100}\right)$ by $15\left(1+\frac{n}{100}\right)$.

When the resolution was decreased by $n$ percent, the new resolution was $75\left(1-\frac{n}{100}\right)$.

(Note that $n$ cannot be larger than 100, since the resolution cannot be decreased by more than $100 \%$.)

Therefore, the number of pixels in the new image is

$$
\left[10\left(1+\frac{n}{100}\right) \times 75\left(1-\frac{n}{100}\right)\right] \times\left[15\left(1+\frac{n}{100}\right) \times 75\left(1-\frac{n}{100}\right)\right]
$$

Since we know that the number of pixels in the new image is 345600 , then

$$
\begin{aligned}
{\left[10\left(1+\frac{n}{100}\right) \times 75\left(1-\frac{n}{100}\right)\right] \times\left[15\left(1+\frac{n}{100}\right) \times 75\left(1-\frac{n}{100}\right)\right] } & =345600 \\
{[10 \times 75] \times[15 \times 75] \times\left(1+\frac{n}{100}\right)^{2} \times\left(1-\frac{n}{100}\right)^{2} } & =345600 \\
843750\left(1+\frac{n}{100}\right)^{2}\left(1-\frac{n}{100}\right)^{2} & =345600 \\
\left(1-\frac{n^{2}}{100^{2}}\right)^{2} & =0.4096 \\
1-\frac{n^{2}}{100^{2}} & = \pm 0.64 \\
1-\frac{n^{2}}{100^{2}} & =0.64 \\
\frac{n^{2}}{100^{2}} & =0.36 \\
\frac{n}{100} & =0.6 \\
n & =60
\end{aligned}
$$

$$
\begin{array}{rlrl}
1-\frac{n^{2}}{100^{2}} & =0.64 & & (n \text { cannot be larger than } 100) \\
\frac{n^{2}}{100^{2}} & =0.36 & \\
\frac{n}{100} & =0.6 & & (\text { since } n \text { must be positive })
\end{array}
$$

Thus, $n=60$.

Final answer: 60