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 \(k\).

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