CEMC Banner

Problem of the Week
Problem C and Solution
Cycles of Eclipses


A planet in a distant solar system has a moon and a sun. On this planet, there is a total solar eclipse whenever the following is true.

On this planet, there is a full moon every \(16\) days. Also, every \(12\) days, the moon is at its closest point to the planet. As well, every \(n\) days the centre of the moon is in line with the centres of the planet and the sun.

If \(n\) is greater than \(10\) but less than \(20\), and total solar eclipses happen on this planet every \(240\) days, determine the value of \(n\).


Since total solar eclipses happen every \(240\) days on this planet, it follows that \(240\) is the least common multiple (LCM) of \(16\), \(12\), and \(n\).

To determine the value of \(n\), we will rewrite each of \(16\), \(12\), and \(240\) as a product of prime numbers. This is known as prime factorization. \[\begin{aligned} 16 &= 2 \times 2 \times 2 \times 2\\ 12 &= 2 \times 2 \times 3\\ 240 &= 2 \times 2 \times 2 \times 2 \times 3 \times 5 \end{aligned}\] The LCM is calculated by determining the greatest number of each prime number in any of the factorizations, and then multiplying these numbers together. From the prime factorizations of \(16\) and \(12\), we can determine that their LCM is equal to \(2 \times 2 \times 2 \times 2 \times 3 =48\). Since \(240\) has an extra factor of \(5\), and \(240\) is the LCM of \(16\), \(12\), and \(n\), it follows that \(5\) must be a factor of \(n\). The only number with a factor of \(5\) that is greater than \(10\) but less than \(20\) is \(15\).

Since the prime factorization of \(15\) is \(15=3 \times 5\), we can conclude that the LCM of \(16\), \(12\), and \(15\) is \(240\), as desired. Therefore \(n=15\).

Extension: Research what conditions must occur for there to be a total solar eclipse on Earth. How often do total solar eclipses occur on Earth?