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Problem of the Week
Problem E and Solution
Looking for Integers


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?


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 \(1+5+9+13+17+21=66\).