Numerical properties of 191

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Determine the numerical properties of 191

We start by listing out divisors for 191
DivisorDivisor Math
1191 ÷ 1 = 191

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)

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 digitsm = 13 + 93 + 13
Square sum of digitsm = 1 + 729 + 1
Square sum of digitsm = 731
Since 731 <> 191, 191 is NOT narcissistic (plus perfect)

Catalan Test:
The nth Catalan number Cn is denoted by:
Cn  =  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)!

Using our factorial lesson to evaluate, we get
C6  =  12!
  6!7!

C6  =  479001600
  (720)(5040)

C6  =  479001600
  3628800

C6 = 132
Since this does not equal 191, this is NOT a Catalan number

Property Summary for the number 191
  ·  Whole
  ·  Prime
  ·  Deficient
  ·  Odd
  ·  Odious
  ·  Undulating
  ·  Palindrome
  ·  Palindromic Prime

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