Base | Representation |
---|---|
bin | 1011101001011100101011… |
… | …10111001011001101001011 |
3 | 10100200121200011220101200221 |
4 | 11310232111313023031023 |
5 | 11324122222442130201 |
6 | 130250345132410511 |
7 | 5252336014642624 |
oct | 564562567131513 |
9 | 110617604811627 |
10 | 25613404255051 |
11 | 8185646a64328 |
12 | 2a58071316a37 |
13 | 113a445078462 |
14 | 6479a41a4a4b |
15 | 2e63e40cb4a1 |
hex | 174b95dcb34b |
25613404255051 has 2 divisors, whose sum is σ = 25613404255052. Its totient is φ = 25613404255050.
The previous prime is 25613404255009. The next prime is 25613404255067. The reversal of 25613404255051 is 15055240431652.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 25613404255051 - 29 = 25613404254539 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 25613404254989 and 25613404255007.
It is not a weakly prime, because it can be changed into another prime (25613404258051) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 12806702127525 + 12806702127526.
It is an arithmetic number, because the mean of its divisors is an integer number (12806702127526).
Almost surely, 225613404255051 is an apocalyptic number.
25613404255051 is a deficient number, since it is larger than the sum of its proper divisors (1).
25613404255051 is an equidigital number, since it uses as much as digits as its factorization.
25613404255051 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 720000, while the sum is 43.
The spelling of 25613404255051 in words is "twenty-five trillion, six hundred thirteen billion, four hundred four million, two hundred fifty-five thousand, fifty-one".
• e-mail: info -at- numbersaplenty.com • Privacy notice • done in 0.071 sec. • engine limits •