Be kind for 2020 (bkf2020)'s Math log

I managed to find that 3² and 7⁵ divide the final number, but only knew that 5⁴, 2 × 5⁴, or 4 × 5⁴ divided the final number. I didn't notice to take 149ⁿ - 2ⁿ mod 5 to then prove that 4 divides the final number.

Another mistake I made was writing 7⁶ in the final number instead of 7⁵ for some reason...

Let O be the center of the square with label 2000 and ABCD be the square with label 1099. Let P be the center of the square with label 200. Then one of these two cases can occur:

			P
	A--B
	|  |
	C--D
O
					

P
	A--B
	|  |
	C--D
		O
					

Another fact I noticed was that ceil(1099/n) gave the row the square with label 1099 should be in, and this may be used with the first fact to calculate the slopes, but that would be very messy. Need to try more.

I thought of n being the multiples of φ(3^3), φ(5^5), and φ(7^7). (See Euler's Totient function.) However, n can be smaller, as shown by manual calculation for 3^3.

To find the value of n so 2^n was congruent to 149^n mod 3^3 = 27, I found the value of n so 2^n was congruent to 14^n mod 27. I then multiplied by the inverse of 2^n (since the inverse of 2 exists mod 27) to get 7^n congruent to 1 mod 27. The smallest value of n should divide φ(3^3) = 18, and I found it to be 9. So n should be a multiple of 9.

For 5^5 and 7^7, I thought of finding the inverses, but that isn't as easy.

1. it assumed the people sitting together have to be adjacent, and
2. the people sitting together have a fixed position, but they can move around.

My original solution was wrong, because it assumed 4 would always have a modulo inverse mod n + 5, but this may not always be true. So, I glanced at MAA solution 1 and completed a similiar solution.

But, this new approach was wrong because it forgot to test the values of n. Just because n^2(n+1)^2 leaves a remainder of 68 mod n + 5 does NOT mean n^2(n+1)^2/4 leaves a remainder of 17 mod n + 5. Also, I forgot that we needed n + 5 > 17, but this did not matter here, and checking the values of n would be an additional measure against this mistake.

Lastly, checking the values of n wasn't as tedious as it looked, but I should be careful here, since the computation isn't super trivial.

Leaving it as 'started' because I could have thought more.

Avoided getting stuck on one approach for this problem. I believe I noticed that I wasn't using the fact that the entire P(x) polynomial was given, so I wrote

• P(x) = x^2 - 3x - 7
• Q(x) = x^2 - ax + 2
• R(x) = x^2 - bx + c

Then, I used this fact from the problem:

• P + Q = 2(x - p)(x - q)
• P + R = 2(x - p)(x - r)
• Q + R = 2(x - q)(x - r).

After that, I found p + q, pq, p + r, pr, q + r, qr in terms of a, b, c, and so on...

The resulting solution was similiar to Solution 1 on the AoPS wiki.

My solution let P_{i, T} be the probability of getting the desired result if the ant is at a vertex of the top face after i moves AND i - 1 moves. Similarly, let P_{i, B} be the probability of getting the desired result if the ant is at a vertex of the bottom face after i moves AND i - 1 moves.

This meant P_{i, T} = ½ P_{i + 1, T} + ½ P_{i + 2, B}, because the ant can either stay on the top face, or it moves to the bottom face and stays at the bottom face for an additional move. Similarly, P_{i, B} = ½ P_{i + 1, B} + ½ P_{i + 2, T}.

Using this, I calculated the values of P_i,B and P_i,T for 1 ≤ i ≤ 8. Note P_{8, B} = 0 and P_{8, T} = 1.

The answer was 2/3 * P_{1, B} + 1/3 * P_{2, T}, but I got 1/2 * P_{1, B} + 1/2 * P_{2, T}, since I thought the ant only had 2 moves when it started, but it actually has 3 moves.

I realized that P_{i, T} + P_{i, B} = 1 and this can be proved using recursion. This may motivate the idea behind solution 3 on the AoPS wiki. That solution is pretty similar to the solution I outlined here.

First, I was skeptical that ABA₁B₁ was a convex quadrilateral. However, let the intersection of AC and BD be E. Then, angle AEB is (arc AB + arc CD) / 2 while angle BEC is (arc BC + arc AD) / 2. Arc AB is less than arc BC, and arc CD is less than arc AD as AB < BC and CD < AD. This means angle AEB is less than angle BEC. That makes angle AEB acute, and it seems arc AD is less than 180, so triangle AEB is also acute. As a result, ABA₁B₁ is a convex quadrilateral.

In the future, I should draw the circle first and then draw ABCD.

After that, I wasn't so sure on where to continue, so I looked at Solution 1 and Solution 1.1. I didn't know how to come up with such a solution, so I tried coming up with my own solution. I did some angle chasing and got where you would come up with solution 1. The mistake I made was getting triangle EC₁D₁ similiar to triangle EDC instead of triangle EC₁D₁ similiar to triangle ECD. Then, I understood where you would come up with solution 1.1 by eliminating A₁, B₁, C₁, and D₁ and trying law of cosines since the absolute values of the cosines of angles AEB, BEC, CED, and AED are the same.

Previously: Tried solution 2 as it was similiar to what I did on the best, but needs lots of computation. Will try to find a cleaner way online. Also may add old problems again as practice.

Currently: Found clean and simple solution. I think the solutions on the wiki are much more complicated than this.

Using a hint from someone, I found a1 > a2 < a3 > a4 < a5 or a1 < a2 > a3 < a4 > a5. However, I got the wrong answer 24. In the first case, I thought that a1, a3, and a5 must be only 3, 4 or 5, and a2 and a4 must be only 1 or 2. In the second case, I thought that a1, a3, and a5 must be only 1, 2, or 3, and a2 and a4 must be only 4 or 5.

However, I did not consider in the first case, a1 = 2, a2 = 1 OR a4 = 1, a5 = 2. In the second case, a1 = 4, a2 = 5 OR a4 = 5, a5 = 4. This adds 8 possible cases.