Base | Representation |
---|---|
bin | 11100100000101010111011… |
… | …010010010001001111000011 |
3 | 121102222012120210001101111021 |
4 | 130200222323102101033003 |
5 | 112413344033321234003 |
6 | 1122403321300330311 |
7 | 35261105301624604 |
oct | 3440527322211703 |
9 | 542865523041437 |
10 | 125390417368003 |
11 | 36a53860756897 |
12 | 120916050a3997 |
13 | 54c735b754808 |
14 | 22d6d0005bdab |
15 | e76a5c3572bd |
hex | 720abb4913c3 |
125390417368003 has 2 divisors, whose sum is σ = 125390417368004. Its totient is φ = 125390417368002.
The previous prime is 125390417367967. The next prime is 125390417368067. The reversal of 125390417368003 is 300863714093521.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 125390417368003 - 29 = 125390417367491 is a prime.
It is not a weakly prime, because it can be changed into another prime (125390417368903) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 62695208684001 + 62695208684002.
It is an arithmetic number, because the mean of its divisors is an integer number (62695208684002).
It is a 1-persistent number, because it is pandigital, but 2⋅125390417368003 = 250780834736006 is not.
Almost surely, 2125390417368003 is an apocalyptic number.
125390417368003 is a deficient number, since it is larger than the sum of its proper divisors (1).
125390417368003 is an equidigital number, since it uses as much as digits as its factorization.
125390417368003 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 3265920, while the sum is 52.
The spelling of 125390417368003 in words is "one hundred twenty-five trillion, three hundred ninety billion, four hundred seventeen million, three hundred sixty-eight thousand, three".
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