Base | Representation |
---|---|
bin | 1011101001011100101110… |
… | …11001110001000010011001 |
3 | 10100200121201222112110110202 |
4 | 11310232113121301002121 |
5 | 11324122301101200441 |
6 | 130250351454351545 |
7 | 5252336450424653 |
oct | 564562731610231 |
9 | 110617658473422 |
10 | 25613430100121 |
11 | 818565a607096 |
12 | 2a58079abb5b5 |
13 | 113a44a5271b6 |
14 | 6479a77b16d3 |
15 | 2e63e64d419b |
hex | 174b97671099 |
25613430100121 has 16 divisors (see below), whose sum is σ = 25906921504128. Its totient is φ = 25320951484800.
The previous prime is 25613430100087. The next prime is 25613430100169. The reversal of 25613430100121 is 12100103431652.
It is not a de Polignac number, because 25613430100121 - 218 = 25613429837977 is a prime.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (25613430108121) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (23) of ones.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 691096931 + ... + 691133991.
It is an arithmetic number, because the mean of its divisors is an integer number (1619182594008).
Almost surely, 225613430100121 is an apocalyptic number.
It is an amenable number.
25613430100121 is a deficient number, since it is larger than the sum of its proper divisors (293491404007).
25613430100121 is a wasteful number, since it uses less digits than its factorization.
25613430100121 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 50512.
The product of its (nonzero) digits is 4320, while the sum is 29.
Adding to 25613430100121 its reverse (12100103431652), we get a palindrome (37713533531773).
The spelling of 25613430100121 in words is "twenty-five trillion, six hundred thirteen billion, four hundred thirty million, one hundred thousand, one hundred twenty-one".
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