Problem C and Solution

Stargazing

In a distant solar system, four different comets: Hypatia, Fibonacci, Lovelace, and Euclid, passed by the planet Ptolemy in \(2023\). On Ptolemy, it is known that the Hypatia comet appears every \(3\) years, the Fibonacci comet appears every \(6\) years, the Lovelace comet appears every \(8\) years, and the Euclid comet appears every \(15\) years.

When is the next year that all four comets will pass by Ptolemy?

Since the Hypatia comet appears every \(3\) years, it will pass by Ptolemy in the following numbers of years: \(3,~6,~9,~12,~15,~18,~21,~24,~27,~30\ldots\).

Since the Fibonacci comet appears every \(6\) years, it will pass by Ptolemy in the following numbers of years: \(6,~12,~18,~24,~30,\ldots\).

Therefore, both the Hypatia and Fibonacci comets will pass by Ptolemy in the following numbers of years: \(6,~12,~18,~24,~30,\ldots\).

This happens because these numbers are *common multiples* of
\(3\) and \(6\). If we want to determine when all four
comets next pass by Ptolemy, we need to find the *least common
multiple* (LCM) of \(3,~6,~8,\) and
\(15\). We shall do this in two
ways.

**Solution 1**

The first way to find the LCM is to list the positive multiples of \(3,~6,~8,\) and \(15\), until we find a common multiple in each list.

Number |
Positive Multiples |
---|---|

\(3\) | \(3,~6,~9,~12,~15,~18,~21,\ldots,~108,~111,~114,~117,~\mathbf{120},~123,\ldots\) |

\(6\) | \(6,~12,~18,~24,~30,~36,~42,\ldots,~96,~102,~108,~114,~\mathbf{120},~126,\ldots\) |

\(8\) | \(8,~16,~24,~32,~40,~48,~56,\ldots,~104,~112,~\mathbf{120},~128,\ldots\) |

\(15\) | \(15,~30,~45,~60,~75,~90,~105,~\mathbf{120},~135,\ldots\) |

Thus, the LCM of \(3,~6,~8,\) and \(15\) is \(120\). Therefore, the next time all four planets will pass by Ptolemy is in \(120\) years. This will be the year \(2143\).

**Solution 2**

The second way to determine the LCM is to rewrite \(3,~6,~8,\) and \(15\) as a prime or a product of prime
numbers. (This is known as *prime factorization*.)

\(3 = 3\)

\(6 = 2 \times 3\)

\(8 = 2 \times 2 \times 2\)

\(15 = 3 \times 5\)

The LCM is calculated by determining the greatest number of each prime number in any of the factorizations (here we will have three \(2\)s, one \(3\), and one \(5\)), and then multiplying these numbers together. This gives \(2 \times 2 \times 2 \times 3 \times 5 = 120\). Therefore, the next time all four planets will pass by Ptolemy is in \(120\) years. This will be the year \(2143\).

Note: The second method is a more efficient way to find the LCM, especially when the numbers are quite large.