11 Jun, 7 tweets, 2 min read
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
2) p = 13; m = ⅒⋅(1 + 3⋅p) = 4;

n=182:
13|182 ⇔ 13|26

n = 6 360 328
13|6 360 328
⇔ 13|636 064
⇔ 13|63 622
⇔ 13|6 370
⇔ 13|637
⇔ 13|91
⇔ 13|13: true

n = 6 360 329
13|6 360 329
⇔ 13|636 068
⇔ 13|63 638
⇔ 13|6 395
⇔ 13|659
⇔ 13|101
⇔ 13|14: false
Corollary:
p | (a + m⋅b) ⇔ p | (a + m⋅b - p⋅b)
⇔ p | (a + (m-p)⋅b)

3) p = 27; m = ⅒⋅(1 + 7⋅p) = 19; m-p = -8

Using m = 19:
27|48 924
⇔ 27|4 968
⇔ 27|648
⇔ 27|216
⇔ 27|135: true

Using m-p = -8
27|48 924
⇔ 27|4 860
⇔ 27|486
⇔ 27|0: true
The corollary is particularly useful when the ones digit of p is 1.

4) p = 41; m = ⅒⋅(1 + 9⋅p) = 37; m-p = -4

41|35 482 261
⇔ 41|3 548 222
⇔ 41|354 814
⇔ 41|35 465
⇔ 41|3 526
⇔ 41|328
⇔ 41|0: true

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# More from @pgeerkens

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.
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.
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.
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/
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/
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:
unless we remove him from this hospital immediately.
I'll help.

The nurse lost her job;
but my Dad lived another 45 years.