# Problem of the Week Problem E and Solution Looking for Integers

## Problem

Suppose $$n$$ and $$k$$ are integers and $$4^k<2024$$. For how many $$(n,k)$$ pairs is $$2024^k \left(\frac{11}{2} \right)^n$$ equal to an integer?

## Solution

First we write $$2024$$ as a product of prime factors: $$2024=2^3 \times 11\times 23$$.

We can then substitute this into our expression. \begin{aligned} 2024^k \left(\frac{11}{2} \right)^n &= \left(2^3 \times 11\times 23 \right)^k \left(\frac{11}{2} \right)^n\\ &= 2^{3k} \times 11^k \times 23^k \times \frac{11^n}{2^n}\\ &= 2^{3k-n} \times 11^{k+n} \times 23^k \end{aligned} Since $$2024^k \left(\frac{11}{2} \right)^n$$ is equal to an integer, it follows that none of the exponents can be negative. Thus, $$3k-n \ge 0$$, $$k+n \ge 0$$, and $$k \ge 0$$.

From $$3k-n \ge 0$$, we can determine that $$n \le 3k$$. Similarly, from $$k+n \ge 0$$, we can determine that $$n \ge -k$$. Thus, $$n$$ is an integer between $$-k$$ and $$3k$$, inclusive.

Since $$4^5=1024$$, $$4^6=4096$$, and $$4^k<2024$$, it follows that $$k \le 5$$. Since $$k \ge 0$$ and $$k$$ is an integer, the possible values of $$k$$ are $$0,~1,~2,~3,~4,$$ and $$5$$.

In the table below, we summarize the number of values of $$n$$ for each possible value of $$k$$.

$$k$$ Minimum value of $$n$$ Maximum value of $$n$$ Number of values of $$n$$
$$0$$ $$0$$ $$0$$ $$1$$
$$1$$ $$-1$$ $$3$$ $$5$$
$$2$$ $$-2$$ $$6$$ $$9$$
$$3$$ $$-3$$ $$9$$ $$13$$
$$4$$ $$-4$$ $$12$$ $$17$$
$$5$$ $$-5$$ $$15$$ $$21$$

Thus, the total number of $$(n,k)$$ pairs is $$1+5+9+13+17+21=66$$.