How many days will pass before the two ships leave Los Angeles on the same day again?

Question

Two cruise ships leave Los Angeles at the same time. They take 1212 and 88 days, respectively, to reach their destination and return to Los Angeles. The cruise ships each take continuous trips to and from Los Angeles. How many days will pass before the two cruise ships leave Los Angeles on the same day again?

Answer 1

Greetings Alexa from GA,

Thank you for such a great problem,

One way of solving this problem is by considering the relationship between the round trip numbers, and their respective arrival/departure dates.

Assuming the first ship, call it ship one, arrives and then subsequently departs every 1212 days, we get the following arrival/departure schedule:

day 0 - ship one departs on it's first round trip
day 1212 - ship one arrives and then departs on it's second round trip
day 2424 - ship one arrives and then departs on it's third round trip
day 3636 - ship one arrives and then departs on it's fourth round trip
.
.
.
day 1212 * m, (where m is some integer) - ship one arrives and then departs on it's (m+1)th round trip

Similarly for the second ship

day 0 - ship one departs on it's first round trip
day 88 - ship one arrives and then departs on it's second round trip
day 176 - ship one arrives and then departs on it's third round trip
day 264 - ship one arrives and then departs on it's fourth round trip
.
.
.
day 88 * n, (where n is some integer) - ship one arrives and then departs on it's (n+1)th round trip

So what we need to find are the smallest trip numbers m and n, that depart on the same day, or

1212 * m = 88 * n.

Since we are only considering full round trips, m and n must be integers and cannot be fractions. So to summarize, in order to find the shortest number of days that need to pass before the two cruise ships leave on the same day after day zero, we need to find the smallest integers m>0 and n>0 that satisfy the above relationship.

The easiest way to find the smallest such a pair of integers, is to first factor the coefficients of this equation into their prime factors, and the remove the common prime factors.
For example in our case we have:

1212 = 2*2*3*101 and 88 = 2*2*2*11.

Plugging this into the above we find:

2*2*3*101 * m = 2*2*2*11 * n,

We can see the largest common prime factor is 4=2*2, and what we are left with is:

3 * 101 * m = 2 * 11 * n.

Since there are no more common factors, we can easly see that

m must be = 2 * 11 = 22, and n must be 3 * 101 = 303.

So the earliest possible day is the 26,664th day where ship one is leaving on it's 23rd round trip and ship is leaving on it's 304th.