Base | Representation |
---|---|
bin | 11101010100000100010… |
… | …001011100010101100111 |
3 | 21010120112210022221100012 |
4 | 131110010101130111213 |
5 | 231001003300440312 |
6 | 4141224030040435 |
7 | 265351663620644 |
oct | 35240421342547 |
9 | 7116483287305 |
10 | 2014411343207 |
11 | 70734176228a |
12 | 2864a634911b |
13 | 117c5c1568a3 |
14 | 6d6d8840dcb |
15 | 375ed2b5322 |
hex | 1d50445c567 |
2014411343207 has 2 divisors, whose sum is σ = 2014411343208. Its totient is φ = 2014411343206.
The previous prime is 2014411343173. The next prime is 2014411343209. The reversal of 2014411343207 is 7023431144102.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 2014411343207 - 222 = 2014407148903 is a prime.
Together with 2014411343209, it forms a pair of twin primes.
It is a Chen prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (2014411343209) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (19) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1007205671603 + 1007205671604.
It is an arithmetic number, because the mean of its divisors is an integer number (1007205671604).
Almost surely, 22014411343207 is an apocalyptic number.
2014411343207 is a deficient number, since it is larger than the sum of its proper divisors (1).
2014411343207 is an equidigital number, since it uses as much as digits as its factorization.
2014411343207 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 16128, while the sum is 32.
Adding to 2014411343207 its reverse (7023431144102), we get a palindrome (9037842487309).
The spelling of 2014411343207 in words is "two trillion, fourteen billion, four hundred eleven million, three hundred forty-three thousand, two hundred seven".
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