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.

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)

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

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

@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).

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.

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.

@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/

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/

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.

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.

@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.

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.

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.

@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/

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/

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/

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/

@GWOMaths One might say that physicists study the symmetry of nature, while mathematicians study the nature of symmetry.

1/

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/

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/

3/

@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

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.

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.

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.

From Albert H. Beiler's

Recreations in the Theory of Numbers:

Calculate in your head:

47² = ?

96² = ?

113² = ?

179² = ?

goodreads.com/book/show/5855…

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

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.

a² - b² = (a + b) ⋅ (a - b)

one can obtain by rearrangement

a² = (a + b) ⋅ (a - b) + b².

@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.

consider the situation after Game 1 as the baseline.

@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/

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/

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/

3/

@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.

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.

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.

(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.

@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/

- 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/

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

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

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

⁺ ⁻ ⁽ ⁾

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

@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.

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.

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.

@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, ...}.

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 Lemma 1:

From every distinct partition of n an odd partition of n can be uniquely associated.

From every distinct partition of n an odd partition of n can be uniquely associated.

@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/

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/

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/

3/

@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)

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 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.

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.