Base | Representation |
---|---|
bin | 101010101100011000… |
… | …0100010111010100011 |
3 | 22202122120120000122012 |
4 | 1111120300202322203 |
5 | 3000231424343042 |
6 | 110041354103135 |
7 | 6424000221554 |
oct | 1253060427243 |
9 | 282576500565 |
10 | 91683434147 |
11 | 359789aa088 |
12 | 1592871aaab |
13 | 885181b566 |
14 | 461a71032b |
15 | 25b905be82 |
hex | 1558c22ea3 |
91683434147 has 2 divisors, whose sum is σ = 91683434148. Its totient is φ = 91683434146.
The previous prime is 91683434113. The next prime is 91683434153. The reversal of 91683434147 is 74143438619.
It is a strong prime.
It is an emirp because it is prime and its reverse (74143438619) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 91683434147 - 210 = 91683433123 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 91683434095 and 91683434104.
It is not a weakly prime, because it can be changed into another prime (91683434177) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (17) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 45841717073 + 45841717074.
It is an arithmetic number, because the mean of its divisors is an integer number (45841717074).
Almost surely, 291683434147 is an apocalyptic number.
91683434147 is a deficient number, since it is larger than the sum of its proper divisors (1).
91683434147 is an equidigital number, since it uses as much as digits as its factorization.
91683434147 is an odious number, because the sum of its binary digits is odd.
The product of its digits is 1741824, while the sum is 50.
The spelling of 91683434147 in words is "ninety-one billion, six hundred eighty-three million, four hundred thirty-four thousand, one hundred forty-seven".
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