22 Nov, 12 tweets, 5 min read
@GWOMaths Hint follows
@GWOMaths Rochambeau is just a fancy name for Rock-Paper-Scissors.
@GWOMaths If you're having difficulty visualizing this,
consider the situation after Game 1 as the baseline.

1) How many ways are there of getting back to that baseline situation?

2) What occurs to the combined number of wins by Alana and Beth for each possibility?
@GWOMaths Solution follows
@GWOMaths After the first game:
Alana/Beth have accumulated one win against each other,
resulting in the winner staying in the arena
while the other retires to be replaced by Chen;
with one Alana v Beth game played.

AB_Games (G) = 1
AB_Wins (W) = 1
@GWOMaths Now Chen can win any number of games,
each of which has no affect on our two tracking stats above
but merely alternates which of Alana/Beth is in the arena and which is on the side-lines.

We still have
AB_Games (G) = 1
AB_Wins (W) = 1
throughout while Chen's win streak holds.
@GWOMaths Then Chen loses, and the following happens:
- AB_Wins increments to 2 (the win over Chen);
- Alana & Beth have another (their second) head-to-head match, so AB_Games increments to 2;
- AB_Wins increments to 3 (the win over the other);
- One of Alana/Beth retires on the loss.
@GWOMaths We are now back to an arena with Chen and either Alana or Beth, with our tracking stats at:

AB_Games (G) = 2
AB_Wins (W) = 3

This cycle repeats every time Chen loses, with
AB_Games incremented once and
AB_Wins incremented twice
@GWOMaths The pattern is that at all time the invariant
AB_Wins + 1 = 2 * AB_Games
holds:

After Game 1:
AB_Games (G) =1
AB_Wins (W) = 1

After Chen Loss #1
AB_Games (G) = 2
AB_Wins (W) = 3

After Chen Loss #2:
AB_Games (G) = 3
AB_Wins (W) = 5
:
:
@GWOMaths Our final situation is that AB_Wins = 20 + 23 = 43;
thus we get by applying our invariant
(viz AB_Wins + 1 = 2 * AB_Games)

AB_Games = (AB_Wins + 1) / 2
= 44 / 2
= 22.

Today's answer is that Alana and Beth played each other
22 times.

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# More from @pgeerkens

6 Nov
@LarrySchweikart My take:
SCOTUS must not be seen to play favourites - yet has an obligation to ensure that corruption is voided.

So wherever widespread and continuous failure to observe due process (as per State law) is seen, void all ballots from that county / counting centre.

1/
@LarrySchweikart If the number of voided counties exceeds more than 1 or 2, then void the entire election for the State.

In the case of House and Senate representatives, require special elections.

2/
@LarrySchweikart For Presidential Electors, legislation now in effect gives the State legislature responsibility and authority to select the State's slate of Electors - as per the original Constitutional design.

3/
20 Mar
@GWOMaths Define
y ≡ x - 16
to allow factoring the term 2¹⁶ completely.

Then 2¹⁶ + 2¹⁹ + 2ˣ
= 2¹⁶ + 2¹⁹ + 2ʸ⁺¹⁶
= 2¹⁶ . (1 + 2³ + 2ʸ)
= 2¹⁶ . (9 + 2ʸ)

Now a solution, viz y=4, is readily obvious as yielding
= 2¹⁶ . (9+16)
= 2¹⁶ . 25
= (2⁸ . 5)²

∴ x = y + 16 = 20.
@GWOMaths However - is our solution unique?

Suppose
∃ a ∈ Z
such that
9 + 2ʸ ≡ a²

Then
2ʸ = a² - 9 = (a-3) . (a+3)
and both (a-3) and (a+3) must be powers of 2.

This only occurs for
a = 5 => 2ʸ = 16 => y = 4 => x = 20.

Our solution is unique.
@GWOMaths Finally - as a fun calculating observation,for those who have memorized powers of two at least to 2¹⁶:

(2⁸ . 5)²
= (2⁷ . 2 . 5)²
= (2⁷ . 10)²
= (128 . 10)²
= 1280²

= 2¹⁶ . 25
= 2¹⁴ . 4 . 25
= 2¹⁴ . 100
= 16,384 . 100
= 1,638,400.
7 Mar
@nklym143 @GWOMaths Then I'd note:

- Generating Function f(x) for a single standard die is
x.(1+x+x²+x³+x⁴+x⁵)
= x+x²+x³+x⁴+x⁵+x⁶
= Σ xⁿ . C(n,1)

1/
@nklym143 @GWOMaths - For two Std. die f(x)² is
= 1.x²+1.x³+1.x⁴+ 1.x⁵+ 1.x⁶+ 1.x⁷
+ 1.x³+1.x⁴+1.x⁵+ 1.x⁶+ 1.x⁷+ 1.x⁸
+ 1.x⁴+1.x⁵+1.x⁶+ 1.x⁷+ 1.x⁸+ 1.x⁹
+ 1.x⁵+1.x⁶+1.x⁷+ 1.x⁸+ 1.x⁹+1.x¹⁰
+ 1.x⁶+1.x⁷+1.x⁸+ 1.x⁹+1.x¹⁰+1.x¹¹
+ 1.x⁷+1.x⁸+1.x⁹+1.x¹⁰+1.x¹¹+1.x¹²

2/
@nklym143 @GWOMaths = 1.x²+2.x³+3.x⁴+4.x⁵+5.x⁶+6.x⁷+5.x⁸+4.x⁹+3.x¹⁰+2.x¹¹+1.x¹²
6 Mar
⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ - supers
Α α - Alpha
Β β - Beta
Γ γ - Gamma
Δ δ - Delta
Ε ε - Epsilon
Ζ ζ - Zeta
Θ θ - Theta
Λ λ -Lambda
Μ μ - Mu
Ν ν - Nu
Π π - Pi
Ρ ρ - Rho
Σ σ - Sigma
Τ τ - Tau
Υ υ - Upsilon
Φ φ - Phi
Χ χ - Chi
Ψ ψ - Psi
Ω ω - Omega

≅ - congruence
∴ - therefore
∈ : Element
Δ : Triangle
± : Plus/minus
× ÷ : Times & Division
≤ ≠ ≥ : Inequality
∠ : Angle
° : Degree
⊥ ∥ : Perpendicular & Parallel
~ : Similarity
≡ : Equivalence
∝ : Proportional to
∞ : Infinity
≪ ≫ : Mush less/greater than
∘ : Function composition
† * : Matrix
More superscripts:
⁺ ⁻ ⁽ ⁾
ᵃ ᵇ ᶜ ᵈ ᵉ ᶠ ᵍ ʰ ⁱ ʲ ᵏ ˡ ᵐ ⁿ ᵒ ᵖ ʳ ˢ ᵗ ᵘ ᵛ ʷ ˣ ʸ ᶻ
ᴬ ᴮ ᴰ ᴱ ᴳ ᴴ ᴵ ᴶ ᴷ ᴸ ᴹ ᴺ ᴼ ᴾ ᴿ ᵀ ᵁ ⱽ ᵂ
ᵝ ᵞ ᵟ ᶿ ᵠ ᵡ
22 Feb
@GWOMaths The game Nim is played by two players alternately removing items from several piles of items. Absent physical materials, with pen and paper a game can be setup as rows of vertical lines, players stroking them out to "remove" them.
@GWOMaths Each turn a player removes 1 or more items from precisely one pile, or row.

The winner is the player who does not (or, as a variant, does) remove the last item.

Consider the setup of two piles of 2 items:

xx
xx

The player to move can be forced to lose by his opponent.
@GWOMaths Note that this is true in both variants.

a) Last item loses:
-Remove one item and your opponent takes both items from the other pile;
- Take both items from one pile and your opponent removes just one from the other.

b) Last item wins: