Determine the numerical properties of 191

We start by listing out divisors for 191

Divisor | Divisor Math |
---|---|

1 | 191 ÷ 1 = 191 |

Whole Numbers are positive numbers, including 0, with no decimal or fractional partsSince 191 ≥ 0 and it is an integer, 191 is a

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

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 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

This can be written as A(191) = Odd

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

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

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

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

A number (n) is automorphic (curious) if n

191

Since 36481 does not end with 191, it is

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

A number (n) is a square if there exists a number m such that m

Analyzing squares, we see that 13

Therefore, 191 is

A number (n) is a cube if there exists a number m such that m

Analyzing cubes, we see that 5

Therefore, 191 is

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

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

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

A number (n) is apocalyptic power if 2

2

Since 2

A pentagonal number is one which satisfies the form:

n(3n - 1) | |

2 |

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

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

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

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

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

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

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

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

n(n + 1)(n + 2) | |

6 |

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

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

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

Square sum of digits

Square sum of digits

Since 731 <> 191, 191 is

The n

C_{n} = | 2n! |

(n + 1)!n! |

Using n = 7, we have:

C_{7} = | (2 x 7)! |

7!(7 + 1)! |

Using our factorial lesson to evaluate, we get

C_{7} = | 14! |

7!8! |

C_{7} = | 87178291200 |

(5040)(40320) |

C_{7} = | 87178291200 |

203212800 |

C

Since this does not equal 191, this is

Using n = 6, we have:

C_{6} = | (2 x 6)! |

6!(6 + 1)! |

Using our factorial lesson to evaluate, we get

C_{6} = | 12! |

6!7! |

C_{6} = | 479001600 |

(720)(5040) |

C_{6} = | 479001600 |

3628800 |

C

Since this does not equal 191, this is

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