P. Geerkens ✝️ Profile picture
Poppy wearer Father of RCAF pilot & 2 beautiful girls Followed by @CoachsCornerDC 🇨🇦 SW; Physics; Math; History Gab/Parler: @pgeerkens
11 Jun
A proof for the Vedic Divisibility Rules discovered by Hindu mathematicians.

For any natural number n = 10a + b with 0 ≤ b ≤ 9;
divisor p coprime to 10; and
M(p) = m as described below

Thm: n is divisible by p exactly when (a+m∙b) is.
M(p) is found from p as:

ones digit of p = 1: m = ⅒⋅(1 + 9⋅p)
ones digit of p = 3: m = ⅒⋅(1 + 3⋅p)
ones digit of p = 7: m = ⅒⋅(1 + 7⋅p)
ones digit of p = 9: m = ⅒⋅(1 + 1⋅p)
Examples:

1) p = 7; m = ⅒⋅(1 + 7⋅p) = 5;

n=84:
a=8, b=4, a+mb = 8+5⋅4 = 28
7|84 ⇔ 7|28

n=32,124,701:
7|32,124,701
⇔ 7|3,212,475
⇔ 7|321,272
⇔ 7|32,137
⇔ 7|3,248
⇔ 7|364
⇔ 7|56
⇔ 7|35
Read 7 tweets
11 Mar
@GWOMaths Spoiler - hint and solution follows
@GWOMaths Define Sₖ(c) as the sum of k consecutive integers with offset c thus:
Sₖ(c) = (c+1) + (c+2) + ... + (c+k-1) + (c+k)
= k⋅c + (1 + 2 + ... + (k-1) + k )
= k⋅c + ½k(k+1).
@GWOMaths We desire to find which n ∈ Z⁺ with 50 ≤ n ≤100 can be expressed as Sₖ(c) for some k ≥ 3 and c ≥ 0.

Case 1 - k is odd:
Sₖ(c)
= k ⋅ (c + ½(k+1))
= k ⋅ ½(2c + k + 1)
where ½(2c + k + 1) is integral.

Case 2 - k is even:
Sₖ(c)
= ½k ⋅ (2c + k+1)
where ½k is integral.
Read 10 tweets
8 Mar
@GWOMaths Two important concepts are:

i) That an expression such as
(5 + √21)
CAN be a perfect square; and

ii) That the resultant square root looks like
(a + b⋅√21)

Then set
(5 + √21)
= (a + b⋅√21)²
= (a² + 21⋅b²) + (2ab √21)

Hence
(1) 5 = a² + 21⋅b²
(2) -1 = 2ab

1/
@GWOMaths From (2):
b = -1 / 2a
and
b² = 1 / 4⋅a².

Substituting into (1);
then rearranging and multiplying through by 4⋅a²:
4⋅a⁴ - 20⋅a² + 21 = 0.

Note that
4⋅21 = 2⋅2⋅3⋅7 = (2⋅3) ⋅ (2⋅7) = 6⋅14
and thus
4⋅a⁴ - 20⋅a² + 21 = (2⋅a² - 3) ⋅ (2⋅a² - 7)

2/
@GWOMaths Hence:
a² = 3/2 or 7/2
a = ±√(3/2) or ±√(7/2)
b = -1 / 2a = ∓1/√6 or ∓1/√14
(a + b⋅√21) = ±( √(3/2) - √(7/2) ) or ±( √(7/2) - √(3/2) )

with those two latter expressions just the sign reversal of each other.

We desire the POSITIVE square root.
Read 6 tweets
10 Feb
@JohnsonFido @capobianco_slv @GWOMaths Got it.

Given
a² - d⋅b² = ±4

we have, equivalently
(½a)² - d⋅(½b)² = ±1

Evaluating:
(½a - √d⋅½b)³
= a⋅½(db²±1) - b⋅½(a²∓1)⋅√d

where both of
(db²±1)
and
(a²∓1)
are always even for d, a, and b odd.

But for d ≅ 1 (mod 4)
then a and b are always the same parity.
@JohnsonFido @capobianco_slv @GWOMaths Thus
x = a⋅½(db²±1)
y = b⋅½(a²∓1)

are positive integer solutions to
x² - d⋅y² = ±1

whenever
a² - d⋅b² = ±4

for positive integers a, b.

I believe this is only relevant for d ≅ 1 (mod 4) even though that condition doesn't appear in the calculation.
Read 4 tweets
17 Jan
@johnrobb We have abandoned competence,
in the futile search for universal expertise.

"What?" I hear you say.
Clearly definitions and explanations are in order.

1/
@johnrobb Expertise, certainly as regards this discussion,
but perhaps also universally, should be regarded
as the knowledge of a very great deal - about very
very little.

2/
@johnrobb To become an expert in a field requires
years - nay, decades - of diligent study of that
solitary field. In diving deeper, the pool in turn
becomes ever smaller: a diving pool rather than a
swimming pool. Nothing is free.

3/
Read 9 tweets
15 Jan
@GWOMaths One might say that physicists study the symmetry of nature, while mathematicians study the nature of symmetry.

1/
@GWOMaths Observing symmetry in nature, such as noting the similarity between the symmetries of a snowflake and a hexagon, is readily comprehensible. What does it mean then to study "the nature of symmetry"?

2/
@GWOMaths Mathematicians define a "group", G, as a set of elements {a,b,c, ...} with a binary operation ⊡ and a distinguished element e (the identity of G) satisfying these specific properties:

3/
Read 14 tweets
2 Dec 20
@pnjaban None of the above:
Field-trained former military paramedic.

I don't want anyone not trained specifically in emergency medicine anywhere near; and none of those civilian trained E.R. either, cause they're trained in CYA first, and I want someone who's serious about saving my life
@pnjaban When my Dad suffered acute hemechromatosis at ~34,
for which the ONLY treatment is *bleeding*,
and the only options "a whole lot" and "a whole lot more",
the attending physician was too embarrassed to prescribe the correct treatment.
@pnjaban The nurse (Irish of course, because they understand hemechromatosis) told me Mom:

Mrs. Geerkens:
Your husband will die tonight
unless we remove him from this hospital immediately.
I'll help.

The nurse lost her job;
but my Dad lived another 45 years.
Read 4 tweets
2 Dec 20
From Albert H. Beiler's
Recreations in the Theory of Numbers:

Calculate in your head:
47² = ?
96² = ?
113² = ?
179² = ?

goodreads.com/book/show/5855…
47² = (47 + 3) ⋅ (47 - 3) + 3² = 50 ⋅ 44 + 9 = 2209

96² = (96+4) ⋅ (96-4) + 4² = 100 ⋅ 92 + 16 = 9216

113² = (113+3) ⋅ (113-3) + 3² = 116⋅110 + 9
= 12760 + 9 = 12,769

179² = (179+21) ⋅ (179-21) + 21²
= 200 ⋅ 158 + (20² + 20 + 21)
= 31600 + 441 = 32041
The trick is from noting that from difference of squares
a² - b² = (a + b) ⋅ (a - b)
one can obtain by rearrangement
a² = (a + b) ⋅ (a - b) + b².
Read 4 tweets
22 Nov 20
@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.
Read 12 tweets
6 Nov 20
@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/
Read 9 tweets
20 Mar 20
@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.
Read 5 tweets
7 Mar 20
@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
Read 11 tweets
6 Mar 20
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
Read 4 tweets
22 Feb 20
@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:
Your opponent wins by mirroring your move.
Read 16 tweets
17 Feb 20
@GWOMaths @MathsTim Hint and then solution follows
@GWOMaths @MathsTim Use modulo arithmetic to narrow the possible values of x and y.
@GWOMaths @MathsTim From inspection, x>=5.
As difference of even numbers is always even, x ≅ 1 mod 2.

Evaluating mod 3, noting 18 ≅ 0:
0² ≅ 0; 0².0 ≅ 0
1² ≅ 1; 1².0 ≅ 0
2² ≅ 1; 2².0 ≅ 0
∴ x x ≅ 1 or 2
Read 10 tweets
15 Feb 20
@GWOMaths @MathsTim Recall from Feb. 1 this about generating functions for the odd and distinct partitions of n.


Then the generating function for unrestricted partitions of n, P(n), is given by

SUM(P(n) . q^n) = PROD(1 / (1 - q^i) )
where n ∈ {0, ... } and i ∈ {1, ...}.
@GWOMaths @MathsTim Then P(7) will be the coefficient of q^7 from the expansion of the RHS, noting that
1 / (1 - q^i)

is just
SUM(1 + q^i + q^2.i + q^3.i + q^4.i + ...)

and that we only require in each expansion powers of q ≤7.
@GWOMaths @MathsTim Se we must find q^7 term in

G
= (1+q+q^2+q^3+q^4+q^5+q^6+q^7)
.(1+q^2+q^4+q^6).(1+q^3+q^6)
.(1+q^4).(1+q^5)(1+q^6).(1+q^7)

Evaluating from right
G
= (1+q+q^2+q^3+q^4+q^5+q^6+q^7)
.(1+q^2+q^4+q^6).(1+q^3+q^6)
.(1+q^4).(1+q^5)(1+q^6+q^7)

...
Read 13 tweets
1 Feb 20
@GWOMaths @MathsTim Lemma 1:
From every distinct partition of n an odd partition of n can be uniquely associated.
@GWOMaths @MathsTim Proof:
For the integer n, any distinct partition can be written as

SUM_i(Gi . 2^Ki)

where every Gi is odd and Ki >= 0

because every positive integer is uniquely expressable as an odd positive integer times a power of 2.
@GWOMaths @MathsTim By grouping on each distinct value of the Gi we get a uniquely associated odd partition of n
Q.E.D.

Lemma 2:
From every odd partition of n a distinct partition of n can be uniquely associated.
Read 8 tweets
29 Jan 20
@GWOMaths @EstherOfReilly Cases n=1,2 trivially false.

Cases n=3,4,5 fail because a friendship graph that is an n-gon fails with friendship # = 2 for all nodes.

1/
@GWOMaths @EstherOfReilly Case n=6:
If the node with highest Friendship # has Friendship # >= 3 then that immediately either defines a triad of mutual friendship or a triad of mutual strangers from the first three friend nodes.

Therefore Friendship # <= 2 for all nodes.

2/
@GWOMaths @EstherOfReilly If the friendship is a single chain we it is have a hexagon, either open or closed, and alternate nodes are mutual strangers.

3/
Read 5 tweets
17 Jan 20
@GWOMaths @MathsTim All "congruence" equations are mod 23.

Expanding gives
2x^2 - 5x -3 ≅ 7

or
2x^2 - 5x - 10 ≅ 0

and equivalently
2x^2 - 5x -10 - 23n = 0 for some small n

To be factored requires that for our n, find two integers with difference
5
and product
2.(-10 -23n)
@GWOMaths @MathsTim Suppose n=0:
Require two integers with difference
5
and product
-20.

None found

Suppose n=1:
Require two integers with difference
5
and product
-66.

6 and -11 work.

Back calculating in
2x^2 - 5x - 10 -23n ≅ 0

gives
2x^2 - 5x - 33 ≅ 0

or
(2x - 11) (x + 3) ≅ 0
@GWOMaths @MathsTim For a +ve solution check the first factor
2x - 11 ≅ 0

means that
2x - 11 = 23 i for some i

i=0 gives no integer solution

i=1 gives x = 17 (as 2.17 = 23 + 11)

So x = 17 is a solution.

Verifying:
2 . 17^2 - 5 . 17 - 3
= 2.289 - 85 - 3
= 578 - 88
= 490
= 21. 23 + 7
≅ 7 mod 23
Read 8 tweets
12 Jan 20
@GWOMaths @MathsTim Don't be intimidated!

This problem is just a straightforward calculation exercise requiring a few definitions.

Those definitions use calculus, but I will give some simple rules to follow to get the calculus right.

This thread may take a few minutes to complete;
bear with me.
@GWOMaths @MathsTim The total derivative (with respect x)
of a polynomial function in one variable x
is the sum of the total derivatives of each term
using this rule

The total derivative of a power function in x is evaluated as:
derivative_wrt_x( c.x^n ) = n.c.x^(n-1)

Evaluate each term & add.
@GWOMaths @MathsTim Note that the total derivative of all constant terms is zero.

This results in another function of x,
of order one less than the original.

Evaluated at each point x,
it represents the slope of the tangent line
to the original function
at that point x.
Read 12 tweets
11 Jan 20
@GWOMaths @MathsTim Hint in next tweet.
@GWOMaths @MathsTim There are many well known "division rules" that enable one to test the divisibility of a number by a small integer much more efficiently than performing the division.

The best known of these is Casting Out Nines:

N is divisible by 9 if and only if the sum of its digits is.
@GWOMaths @MathsTim In most, perhaps all, of these divisible rules the remainder is provided also. This is true for Casting Out Nines, resulting in a Casting Out Threes corollary - more useful to us as 3 is one of the low primes to be tested today.
Read 11 tweets