Base | Representation |
---|---|
bin | 111100100010101110… |
… | …0011110100000000111 |
3 | 110102120220111012022021 |
4 | 1321011130132200013 |
5 | 4112232103034001 |
6 | 135421101344011 |
7 | 12251604500326 |
oct | 1710534364007 |
9 | 412526435267 |
10 | 130014111751 |
11 | 50158638537 |
12 | 21245568007 |
13 | c34caa5a94 |
14 | 64153000bd |
15 | 35ae1eeba1 |
hex | 1e4571e807 |
130014111751 has 2 divisors, whose sum is σ = 130014111752. Its totient is φ = 130014111750.
The previous prime is 130014111739. The next prime is 130014111769. The reversal of 130014111751 is 157111410031.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 130014111751 - 29 = 130014111239 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (130014171751) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65007055875 + 65007055876.
It is an arithmetic number, because the mean of its divisors is an integer number (65007055876).
Almost surely, 2130014111751 is an apocalyptic number.
130014111751 is a deficient number, since it is larger than the sum of its proper divisors (1).
130014111751 is an equidigital number, since it uses as much as digits as its factorization.
130014111751 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 420, while the sum is 25.
Adding to 130014111751 its reverse (157111410031), we get a palindrome (287125521782).
The spelling of 130014111751 in words is "one hundred thirty billion, fourteen million, one hundred eleven thousand, seven hundred fifty-one".
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