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

Divisor Sum = 1

Since our divisor sum of 1 < 191, 191 is

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

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

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

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

191

Since 36481 does not end with 191, it is

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

Analyzing squares, we see that 13

Therefore, 191 is

Analyzing cubes, we see that 5

Therefore, 191 is

The number read backwards is 191

Since 191 is the same backwards and forwards, it is a

From above, since 191 is both prime and a palindrome, it is a

Since there is at least one digit in 191 not equal to 1, then it is

2

Since 2

n(3n - 1) | |

2 |

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

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

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

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

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

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

6 |

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

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

C_{n} = | 2n! |

(n + 1)!n! |

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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