Base | Representation |
---|---|
bin | 1100010010011110010111… |
… | …1110001111100110010111 |
3 | 1202211200121210021112210012 |
4 | 3010213211332033212113 |
5 | 3232333103243221312 |
6 | 44423040021150435 |
7 | 2563114214552264 |
oct | 304474576174627 |
9 | 52750553245705 |
10 | 13511530445207 |
11 | 433a231767353 |
12 | 162276025b41b |
13 | 7701938b5422 |
14 | 349d6675826b |
15 | 1866ec6d8322 |
hex | c49e5f8f997 |
13511530445207 has 2 divisors, whose sum is σ = 13511530445208. Its totient is φ = 13511530445206.
The previous prime is 13511530445123. The next prime is 13511530445209. The reversal of 13511530445207 is 70254403511531.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 13511530445207 - 230 = 13510456703383 is a prime.
It is a super-2 number, since 2×135115304452072 (a number of 27 digits) contains 22 as substring.
Together with 13511530445209, 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 (13511530445209) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6755765222603 + 6755765222604.
It is an arithmetic number, because the mean of its divisors is an integer number (6755765222604).
Almost surely, 213511530445207 is an apocalyptic number.
13511530445207 is a deficient number, since it is larger than the sum of its proper divisors (1).
13511530445207 is an equidigital number, since it uses as much as digits as its factorization.
13511530445207 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 252000, while the sum is 41.
Adding to 13511530445207 its reverse (70254403511531), we get a palindrome (83765933956738).
The spelling of 13511530445207 in words is "thirteen trillion, five hundred eleven billion, five hundred thirty million, four hundred forty-five thousand, two hundred seven".
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