Mathematical Meanderings: cyclic

Thursday, 20 February 2025

An Interesting Triple 7 Number

Today I turned $ \textbf{27717} $ days old and this number has a plethora of interesting properties that deserve a special mention and thus a dedicated post. Here are some of those properties.

$ \textbf{27717} $ is a so-called Lucky Cube, meaning it is a number whose cubes contain the digit sequence “888”, here:$$27717^3 = 21293088810813$$The numbers that satisfy from 27717 to 40000 are: 27717, 27942, 27973, 28192, 28442, 28484, 28692, 28740, 28942, 29079, 29192, 29354, 29387, 29391, 29418, 29420, 29442, 29491, 29642, 29692, 29942, 29989.
$ \textbf{27717} $ is the lesser of a pair of adjacent composite numbers such that both are only one step away from their home primes. Here:

$27717 = 3 \times 9239 \rightarrow 39239$
$27718 = 2 \times 13859 \rightarrow 213859$

$ \textbf{27717} $ is a number such that n + POD(n) and n - POD(n) are both prime (where POD stands for Product Of Digits). Here we have POD = 686:

$27717 + 686 = 28403$ which is a prime number
$27717 - 686 = 27031$ which is a prime number

$ \textbf{27717} $ is an interprime number because it is at equal distance from the previous prime (27701) and the next prime (27733).
$ \textbf{27717} $ is a number whose sum of divisors has prime factors (ignoring multiplicity) that multiply to the factorial 2310 where

$2310= 2 \times 3 \times 5 \times 7 \times 11$

Here 27717 has a sum of divisors 36960 and

$36960= 2^5 \times 3 \times 5 \times 7 \times 11$

but also forms a consecutive pair with 27718 because its sum of the divisors is 41580 and

$41580= 2^2 \times 3^3 \times 5 \times 7 \times11$

See blog post Primorials and the Sigma Function.
$ \textbf{27717} $ is the TENTH member of an interesting number chain (which is base independent):

$27708 = 12 \times 2309$
$27709 = 11 \times 2519$
$27710 = 10 \times 2771$
$27711 = 9 \times 3079$
$27712 = 8 \times 3464$
$27713 = 7 \times 3959$
$27714 = 6 \times 4619$
$27715 = 5 \times 5543$
$27716 = 4 \times 6929$
$27717 = 3 \times 9239$
$27718 = 2 \times 13859$

See blog post Count Down Number Chains

$ \textbf{27717} $ is a cyclic number.
$ \textbf{27717} $ is a xenodrome in base 9 : 42016. See blog post Xenodromes.
$ \textbf{27717} $ is a number that does not reach a palindrome after 2001 cycles of the reverse and add algorithm.
$ \textbf{27717} $ is a D-number meaning it is a number $n > 3$ such that n divides $ k^{n-2}- k$ for all $1 < k < n$ relatively prime to $n$.
$ \textbf{27717} $ can be rendered as a digit equation as follows: $2 - \dfrac{7}{7} = 1 ^ 7$

Tuesday, 12 September 2023

Cyclic Quadrilaterals

I was forced to think about cyclic quadrilaterals when looking for information about the number associated with my diurnal age today. 27190 is a member of OEIS A329950:

A329950

Floor of area of quadrilateral with consecutive prime sides configured as a cyclic quadrilateral.

In the case of 27190, the consecutive prime sides are 157, 163, 167 and 173 and the area is given by Brahmagupta's formula: $$ \text{area }=\sqrt{(s-a) \times (s-b) \times (s-c) \times(s-d)}\\ \text{ where } s=\frac{a+b+c+d}{2} \text{ and }a,b,c \text{ and } d \text{ are the four sides}$$Here the area turns out to be 27190.9834504013 which is 27190 when truncated. The initial members of the sequence are (permalink):

13, 30, 70, 130, 214, 310, 461, 627, 874, 1167, 1423, 1750, 2094, 2512, 2995, 3574, 4137, 4603, 5237, 5829, 6526, 7522, 8507, 9478, 10390, 11014, 11650, 12932, 14314, 16053, 17799, 19278, 20698, 22159, 23994, 25403, 27190, 29033, 30595, 32718, 34558, 36255, 38014, 39954

The radius $R$ of the circumcircle, referred to as the circumradius, is given by the formula:$$R=\frac{1}{4} \sqrt{\frac{(ab+cd)(ac+bd)(ad+bc)}{(s-a)(s-b)(s-c)(s-d)}}$$In the case of the cyclic quadrilateral with sides 157, 163, 167, 173 and truncated area of 27190, the truncated circumradius is 116 and the truncated area of the circumcircle is 42818 square units.

These types of cyclic quadrilaterals are very close to being square in shape. For example, the quadrilateral with sides of 157, 163, 167 and 173 has an average side length of 165 and the area of a square with this side is 27225 square units and thus very close to 27190.

In the formula for the area given earlier, it can be noted that when $d=0$, we get a triangle whose area is given by the familiar Heron's formula:$$ \text{area }=\sqrt{s \times (s-a) \times (s-b) \times (s-c) }\\ \text{ where } s=\frac{a+b+c}{2} \text{ and }\\a,b \text{ and } c \text{ are the three sides of the triangle}$$It should be noted that Brahmagupta's formula does not apply to quadrilaterals in general. It only applies to cyclic quadrilaterals. The more general formula is similar but more complex and I won't cover that here. For more information follow this link.