Base | Representation |
---|---|
bin | 1011000001100111111000… |
… | …0110011011101111101011 |
3 | 1120220220022210111212002022 |
4 | 2300121332012123233223 |
5 | 3042103401402400120 |
6 | 41441001123455055 |
7 | 2360552116052111 |
oct | 260317606335753 |
9 | 46826283455068 |
10 | 12122513325035 |
11 | 3954146814253 |
12 | 143951122748b |
13 | 69c1c04c0519 |
14 | 2dca38a43ab1 |
15 | 160503885025 |
hex | b067e19bbeb |
12122513325035 has 16 divisors (see below), whose sum is σ = 14562116550432. Its totient is φ = 9687946944000.
The previous prime is 12122513325011. The next prime is 12122513325053. The reversal of 12122513325035 is 53052331522121.
It is not a de Polignac number, because 12122513325035 - 214 = 12122513308651 is a prime.
It is a super-2 number, since 2×121225133250352 (a number of 27 digits) contains 22 as substring.
It is a Duffinian number.
It is a junction number, because it is equal to n+sod(n) for n = 12122513324986 and 12122513325004.
It is an unprimeable number.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 29405957 + ... + 29815353.
It is an arithmetic number, because the mean of its divisors is an integer number (910132284402).
Almost surely, 212122513325035 is an apocalyptic number.
12122513325035 is a deficient number, since it is larger than the sum of its proper divisors (2439603225397).
12122513325035 is a wasteful number, since it uses less digits than its factorization.
12122513325035 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 415534.
The product of its (nonzero) digits is 54000, while the sum is 35.
Adding to 12122513325035 its reverse (53052331522121), we get a palindrome (65174844847156).
The spelling of 12122513325035 in words is "twelve trillion, one hundred twenty-two billion, five hundred thirteen million, three hundred twenty-five thousand, thirty-five".
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