Base | Representation |
---|---|
bin | 110000011101011101110100… |
… | …000110101100111100001111 |
3 | 1000221122010122111210212202111 |
4 | 300131131310012230330033 |
5 | 210413420003132342112 |
6 | 2033143025201204451 |
7 | 62615133410136553 |
oct | 6035356406547417 |
9 | 1027563574725674 |
10 | 213131110043407 |
11 | 61a0143a337216 |
12 | 1baa22a1630727 |
13 | 91c023caa802a |
14 | 3a8b85a709863 |
15 | 1999068c565a7 |
hex | c1d7741acf0f |
213131110043407 has 2 divisors, whose sum is σ = 213131110043408. Its totient is φ = 213131110043406.
The previous prime is 213131110043387. The next prime is 213131110043441. The reversal of 213131110043407 is 704340011131312.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 213131110043407 - 227 = 213130975825679 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (213131110043497) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 106565555021703 + 106565555021704.
It is an arithmetic number, because the mean of its divisors is an integer number (106565555021704).
Almost surely, 2213131110043407 is an apocalyptic number.
213131110043407 is a deficient number, since it is larger than the sum of its proper divisors (1).
213131110043407 is an equidigital number, since it uses as much as digits as its factorization.
213131110043407 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 6048, while the sum is 31.
Adding to 213131110043407 its reverse (704340011131312), we get a palindrome (917471121174719).
The spelling of 213131110043407 in words is "two hundred thirteen trillion, one hundred thirty-one billion, one hundred ten million, forty-three thousand, four hundred seven".
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