Base | Representation |
---|---|
bin | 111100101101001111… |
… | …1110101111011000011 |
3 | 110110111112002210222012 |
4 | 1321122133311323003 |
5 | 4113443013100101 |
6 | 135520120003135 |
7 | 12263423620316 |
oct | 1713237657303 |
9 | 413445083865 |
10 | 130367315651 |
11 | 50319a50891 |
12 | 213239004ab |
13 | c3a8011381 |
14 | 644a1a297d |
15 | 35d021cbbb |
hex | 1e5a7f5ec3 |
130367315651 has 2 divisors, whose sum is σ = 130367315652. Its totient is φ = 130367315650.
The previous prime is 130367315627. The next prime is 130367315653. The reversal of 130367315651 is 156513763031.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 130367315651 - 214 = 130367299267 is a prime.
Together with 130367315653, it forms a pair of twin primes.
It is a Chen prime.
It is a junction number, because it is equal to n+sod(n) for n = 130367315599 and 130367315608.
It is not a weakly prime, because it can be changed into another prime (130367315653) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65183657825 + 65183657826.
It is an arithmetic number, because the mean of its divisors is an integer number (65183657826).
Almost surely, 2130367315651 is an apocalyptic number.
130367315651 is a deficient number, since it is larger than the sum of its proper divisors (1).
130367315651 is an equidigital number, since it uses as much as digits as its factorization.
130367315651 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 170100, while the sum is 41.
The spelling of 130367315651 in words is "one hundred thirty billion, three hundred sixty-seven million, three hundred fifteen thousand, six hundred fifty-one".
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