Show numerical properties of 191
We start by listing out divisors for 191
Divisor | Divisor Math |
---|---|
1 | 191 ÷ 1 = 191 |
Positive Numbers > 0
Since 191 ≥ 0 and it is an integer
191 is a positive number
Positive numbers including 0
with no decimal or fractions
Since 191 ≥ 0 and it is an integer
191 is a whole number
Since 191 is only divisible by 1 and itself
it is a prime number
Calculate divisor sum D
If D = N, then it's perfect
If D > N, then it's abundant
If D < N, then it's deficient
Divisor Sum = 1
Since our divisor sum of 1 < 191
191 is a deficient number!
A number is even if it is divisible by 2
If not divisible by 2, it is odd
95.5 = | 191 |
2 |
Since 95.5 is not an integer, 191 is not divisible by
it is an odd number
This can be written as A(191) = Odd
Get binary expansion
If binary has even amount 1's, then it's evil
If binary has odd amount 1's, then it's odious
191 to binary = 10111111
There are 7 1's, 191 is an odious number
Can you stack numbers in a pyramid?
Each row above has one item less than the row before it
Using a bottom row of 20 items, we cannot form a pyramid
191 is not triangular
Is there an integer m such that n = m(m + 1)
No integer m exists such that m(m + 1) = 191
191 is not rectangular
Does n2 ends with n
1912 = 191 x 191 = 36481
Since 36481 does not end with 191
it is not automorphic (curious)
Do the digits of 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
Is there a number m such that m2 = n?
132 = 169 and 142 = 196 which do not equal 191
Therefore, 191 is not a square
Is there a number m such that m3 = n
53 = 125 and 63 = 216 ≠ 191
Therefore, 191 is not a cube
Is the number read backwards equal to the number?
The number read backwards is 191
Since 191 is the same backwards and forwards
it is a palindrome
Is it both prime and a palindrome
From above, since 191 is both prime and a palindrome
it is a palindromic prime
A number is repunit if every digit is equal to 1
Since there is at least one digit in 191 ≠ 1
then it is NOT repunit
Does 2n contain the consecutive digits 666?
2191 = 3.1385508676933E+57
Since 2191 does not have 666
191 is NOT an apocalyptic power
It satisfies the form:
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 NOT a pentagonal number
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
Is there an integer m such that n = m(2m - 1)
No integer m exists such that m(2m - 1) = 191
Therefore 191 is not hexagonal
Is there 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
Is there an integer m such that n = m(3m - 3)
No integer m exists such that m(3m - 2) = 191
Therefore 191 is not octagonal
Is there 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
Tetrahederal numbers satisfy the form:
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 NOT a tetrahedral (Pyramidal) number
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
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)
Cn = | 2n! |
(n + 1)!n! |
C7 = | (2 x 7)! |
7!(7 + 1)! |
Using our factorial lesson
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
C6 = | (2 x 6)! |
6!(6 + 1)! |
Using our factorial lesson
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