The other day I tried to answer a question from the website brilliant.org and although I got the correct answer through trial and error I didn't find a nice solution to the question, and I want to see if there is any nice way to get the answer using trigonometry or another method.

On the planet Blue Ox, there are three moons: a red one, a blue one, and a regular one. The red moon orbits around the planet once every 66 days, the blue moon orbits around the planet every 3 days, and the regular moon orbits around the planet every 30 days. The inhabitants of Blue Ox love to throw pies at newcomers. They would usually throw their pies at the newcomer unless that the day is when their three moons line up in a row, because they think that day is unlucky. Josh the Astronaut has just arrived on Blue Ox. The probability that he would be hit with pies is the fraction a/b (a divided by b). What is a+b?

Edited 3/15/2014 13:23:50

First of all, he'll be hit by pies every day except for one day (As if it is two or more days, we can divide both numbers by the number of days he doesn't get hit).

So a = b-1.

Now what we need to do is find the lowest common multiple of 66, 30, and 3. This can be done by taking the prime factors of 66, 30, and 3.

66: 2, 3, 11
30: 2, 3, 5
03: 3

Now we take each factor that appears in at least one number. (Note: If there were duplicates of primes in one number, such as "8: 2, 2, 2" you have to take all of them. In this case, there are none.)*

2*3*5*11 = 330 = b

a = 330-1
a = 329

329/330 chance of being hit
a + b = 329 + 330 = 659

Of course this assumes that all planets only line up at one point in rotation - it might be possible that they line up three times. To prevent this, I propose that the Blue Moon orbits clockwise, the Red Moon counterclockwise, and the Normal Moon whichever way is normal.

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*Every number has a unique prime factorization. If number X is a multiple of number Y, then number X will have all the factors of number Y. By making sure all of the factors of 66, 30, and 3 are in that number, the number created will be a multiple of all three numbers. By using prime factors, it also ensures that the number created will be the lowest common multiple.

Edited 3/15/2014 15:25:07

Of course this assumes that all planets only line up at one point in rotation - it might be possible that they line up three times. To prevent this, I propose that the Blue Moon orbits clockwise, the Red Moon counterclockwise, and the Normal Moon whichever way is normal.

Sorry, forgot to add that in the start, lets assume they all travel clockwise. I was looking for a good way to prove they line up 3 times in the time period 330 days.

In that case we have to take the frequency of all three moons. This is done by taking the reciprocal of the period (how long it takes to make one orbit.)

How frequency works: 1(revolution)/3 (days) - It makes one revolution every 3 days.

In this case, X refers to days. Multiplying frequency by days just gives us revolutions, which is what we need! Since we don't care about how many full revolutions it makes, but only the position its in with its current revolution, we take mod 1.*

If ((1/3)X)mod 1 = ((1/30)X)mod 1 = ((1/66)X)mod 1 = Y
Then the three moons are in the same place.
We'll need to limit X, as otherwise this equation has infinite answers. Since the period is 330 days, we can limit it as such: 0 < X ≤ 330.

Now, obviously, checking every single day up until 330 is quite tedious. But, we can reduce this: The only possible times all three can meet are if Y = 0, 1/3, or 2/3, because the first equation can only produce those three numbers. Thus, the second equation can only possibly equal the first every 10 days, while the third equation can only possibly equal the first every 11 days.
Since 10 and 11's LCM is 110, we only have to check 110, 220, and 330. All three work.

*1 full revolution takes the moon back to where it started, which is why we don't need it. Since revolutions are actually an angle (2pi radians/360 degrees), it doesn't matter if these moons have different sized orbits.

Something a bit harder this time:

In this question t(A) is part of the sequence stated below, with position in the sequence A. t(A+1) is the next number in the sequence.

A sequence of numbers t0,t1,t2,...satisfies t(n+2) = p*t(n+1) + q*t(n) (n greater than or equal to 0), where p and q are real. Throughout this question, x, y and z are non-zero real numbers.

(i) Show that, if t(n) = x for all values of n, then p + q = 1 and x can be any (non-zero) real number.

(ii) Show that, if t(2n) = x, and t(2n+1) = y, for all values of n, then q ± p = 1. Deduce that either x = y or x = −y, unless p and q take certain values that you should identify.

(iii) Show that, if t(3n) = x,t(3n+1) = y and t(3n+2) = z for all values of n, then p^3 + q^3 + 3pq − 1 = 0. Deduce that either p + q = 1 or (p−q)^2 + (p+1)^2 + (q+1)^2 = 0. Hence show that either x = y = z or x + y + z = 0

i) if t(n)=x for any n --> t(n+1)=x and t(n+2)=x
We apply the first fórmula:
t(n+2) = p*t(n+1) + q*t(n) --> x = p*x + q*x = (p+q)*x --> 1 = p+q (we can divide by x because x!=0)//

ii) t(2n) = x, and t(2n+1) = y, for all values of n ==> n even --> x, n odd --> y
Then the t() function values are (starting in n=0):
x, y, x, y, x, y, x, y, x, ...

t(n+2) = p*t(n+1) + q*t(n) --> x = py +qx (n even)
t(n+2) = p*t(n+1) + q*t(n) --> y = px +qy (n odd)

Equation System:
y = px +qy
x = py +qx

We can transform them in:
x = (1-q)y/p
x = py/(1-q)

Then:
(1-q)y/p = py/(1-q) --> (1-q)/p = p/(1-q) --> (1-q)^2 = p^2 --> 1-q = ±p -->q ± p = 1

q= 1± p then:
y = px +qy = px +(1± p)y --> 0 = px ± py --> x±y = 0 --> x=-y or x=y ONLY if p!=0 (and thus q=1)

iii) Didn't try it but it's similar to last one. The serie is:
x, y, z, x, y, z, x, y, z, x, ...

E.System:
x=pz+qy
y=px+qz
z=py+qx

Try to complete this alone, it's mostly the same so I think you will be able to do it ;)

Edited 3/15/2014 21:59:51