Numerical properties of 191

Determine the numerical properties of 191

We start by listing out divisors for 191

Divisor	Divisor Math
1	191 ÷ 1 = 191

Positive or Negative Number Test:

Positive Numbers are greater than 0

Since 191 ≥ 0 and it is an integer, 191 is a positive number

Whole Number Test:

Whole Numbers are positive numbers, including 0, with no decimal or fractional parts

Since 191 ≥ 0 and it is an integer, 191 is a whole number

Prime or Composite Test:

Since 191 is only divisible by 1 and itself, it is a prime number

Perfect/Deficient/Abundant Test:

A perfect number is a number who has a divisor sum equal to itself. An abundant number is a number who has a divisor sum greater than the number, otherwise, it is deficient.
Divisor Sum = 1
Since our divisor sum of 1 < 191, 191 is a deficient number!

Odd or Even Test (Parity Function):

A number is even if it is divisible by 2, else it is odd

95.5 =	191
	2

Since 95.5 is not an integer, 191 is not divisible by 2, and therefore, it is an odd number
This can be written as A(191) = Odd

Evil or Odious Test:

A number is evil is there are an even number of 1's in the binary expansion, else, it is odious.
Using our decimal to binary calculator, we see the binary expansion of 191 is 10111111
Since there are 7 1's in the binary expansion which is an odd number, 191 is an odious number

Triangular Test:

A number is triangular if it can be stacked in a pyramid with each row above containing one item less than the row before it, ending with 1 item at the top
Using a bottom row of 20 items, we cannot form a pyramid with our numbers, therefore 191 is not triangular

Triangular number: 1st 2nd 3rd 4th 5th 6th 7th 8th 9th 10th

Rectangular Test:

A number n is rectangular is if there is an integer m such that n = m(m + 1)
No integer m exists such that m(m + 1) = 191, therefore 191 is not rectangular

Rectangular number: 1st 2nd 3rd 4th 5th 6th 7th 8th 9th 10th

Automorphic (Curious) Test:

A number (n) is automorphic (curious) if n² ends with n
191² = 191 x 191 = 36481
Since 36481 does not end with 191, it is not automorphic (curious)

Automorphic number: 1st 2nd 3rd 4th 5th 6th 7th 8th 9th 10th

Undulating Test:

A number (n) is undulating if the digits that comprise n alternate in the form abab
In this case, a = 1 and b = 9
In order to be undulating, Digit 1: 111 should be equal to 1
In order to be undulating, Digit 2: 999 should be equal to 9
In order to be undulating, Digit 3: 111 should be equal to 1
Since all 3 digits form our abab undulation pattern, 191 is undulating

Square Test:

A number (n) is a square if there exists a number m such that m² = n
Analyzing squares, we see that 13² = 169 and 14² = 196 which do not equal 191
Therefore, 191 is not a square

Cube Test:

A number (n) is a cube if there exists a number m such that m³ = n
Analyzing cubes, we see that 5³ = 125 and 6³ = 216 which do not equal 191
Therefore, 191 is not a cube

Palindrome Test:

A number (n) is a palindrome if the number read backwards equals the number itself
The number read backwards is 191
Since 191 is the same backwards and forwards, it is a palindrome

Palindromic Prime Test:

A number is a palindromic prime if it is both prime and a palindrome
From above, since 191 is both prime and a palindrome, it is a palindromic prime

Repunit Test:

A number is repunit if every digit is equal to 1
Since there is at least one digit in 191 not equal to 1, then it is NOT repunit

Apocalyptic Power Test:

A number (n) is apocalyptic power if 2ⁿ contains the consecutive digits 666.
2¹⁹¹ = 3.1385508676933E+57
Since 2¹⁹¹ does not contain the sequence 666, 191 is NOT an apocalyptic power

Pentagonal Test:

A pentagonal number is one which satisfies the form:

n(3n - 1)
2

Check values of 11 and 12

Using n = 12, we have:

12(3(12 - 1)
2

12(36 - 1)
2

12(35)
2

420
2

210 ← Since this does not equal 191, this is NOT a pentagonal number

Using n = 11, we have:

11(3(11 - 1)
2

11(33 - 1)
2

11(32)
2

352
2

176 ← Since this does not equal 191, this is NOT a pentagonal number

Pentagonal number: 1st 2nd 3rd 4th 5th 6th 7th 8th 9th 10th

Hexagonal Test:

A number n is hexagonal is if there is an integer m such that n = m(2m - 1)
No integer m exists such that m(2m - 1) = 191, therefore 191 is not hexagonal

Hexagonal number: 1st 2nd 3rd 4th 5th 6th 7th 8th 9th 10th

Heptagonal Test:

A number n is heptagonal is if there is an integer m such that:

m =	n(5n - 3)
	2

No integer m exists such that m(5m - 3)/2 = 191, therefore 191 is not heptagonal

Heptagonal number: 1st 2nd 3rd 4th 5th 6th 7th 8th 9th 10th

Octagonal Test:

A number n is octagonal is if there is an integer m such that n = m(3m - 3)
No integer m exists such that m(3m - 2) = 191, therefore 191 is not octagonal

Octagonal number: 1st 2nd 3rd 4th 5th 6th 7th 8th 9th 10th

Nonagonal Test:

A number n is nonagonal is if there is an integer m such that:

m =	n(7n - 5)
	2

No integer m exists such that m(7m - 5)/2 = 191, therefore 191 is not nonagonal

Nonagonal number: 1st 2nd 3rd 4th 5th 6th 7th 8th 9th 10th

Tetrahedral (Pyramidal) Test:

A tetrahedral (Pyramidal) number is one which satisfies the form:

n(n + 1)(n + 2)
6

Check values of 9 and 10

Using n = 10, we have:

10(10 + 1)(10 + 2)
6

10(11)(12)
6

1320
6

220 ← Since this does not equal 191, this is NOT a tetrahedral (Pyramidal) number

Using n = 9, we have:

9(9 + 1)(9 + 2)
6

9(10)(11)
6

990
6

165 ← Since this does not equal 191, this is NOT a tetrahedral (Pyramidal) number

Narcissistic (Plus Perfect) Test:

An m digit number n is narcissistic if it is equal to the square sum of it's m-th powers of its digits
191 is a 3 digit number, so m = 3
Square sum of digits^m = 1³ + 9³ + 1³
Square sum of digits^m = 1 + 729 + 1
Square sum of digits^m = 731
Since 731 <> 191, 191 is NOT narcissistic (plus perfect)

Catalan Test:

The n^th Catalan number C_n is denoted by:

C_n =	2n!
	(n + 1)!n!

Check values of 6 and 7

Using n = 7, we have:

C₇ =	(2 x 7)!
	7!(7 + 1)!

Using our factorial lesson to evaluate, we get

C₇ =	14!
	7!8!

C₇ =	87178291200
	(5040)(40320)

C₇ =	87178291200
	203212800

C₇ = 429
Since this does not equal 191, this is NOT a Catalan number

Using n = 6, we have:

C₆ =	(2 x 6)!
	6!(6 + 1)!