Base | Representation |
---|---|
bin | 11010000010111111110110… |
… | …100110011001100101001011 |
3 | 120000121101000112211110011011 |
4 | 122002333312212121211023 |
5 | 110003333032230404003 |
6 | 1043345534215144351 |
7 | 33062224616002354 |
oct | 3202776646314513 |
9 | 500541015743134 |
10 | 114555210013003 |
11 | 33556656671181 |
12 | 10a216980750b7 |
13 | 4bbc682497cb6 |
14 | 2040704b22d2b |
15 | d39c9d28346d |
hex | 682ff699994b |
114555210013003 has 2 divisors, whose sum is σ = 114555210013004. Its totient is φ = 114555210013002.
The previous prime is 114555210012979. The next prime is 114555210013021. The reversal of 114555210013003 is 300310012555411.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-114555210013003 is a prime.
It is a super-2 number, since 2×1145552100130032 (a number of 29 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (114555210013063) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 57277605006501 + 57277605006502.
It is an arithmetic number, because the mean of its divisors is an integer number (57277605006502).
Almost surely, 2114555210013003 is an apocalyptic number.
114555210013003 is a deficient number, since it is larger than the sum of its proper divisors (1).
114555210013003 is an equidigital number, since it uses as much as digits as its factorization.
114555210013003 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 9000, while the sum is 31.
Adding to 114555210013003 its reverse (300310012555411), we get a palindrome (414865222568414).
The spelling of 114555210013003 in words is "one hundred fourteen trillion, five hundred fifty-five billion, two hundred ten million, thirteen thousand, three".
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