Base | Representation |
---|---|
bin | 100010101011110000100000… |
… | …100001011010010001110001 |
3 | 202000002201221222100001210110 |
4 | 202223300200201122101301 |
5 | 124443211124123122113 |
6 | 1300232104111251533 |
7 | 44062460241334242 |
oct | 4253604041322161 |
9 | 660081858301713 |
10 | 152540604114033 |
11 | 4467111a32a383 |
12 | 15137483819ba9 |
13 | 6716695509915 |
14 | 299500c9d4cc9 |
15 | 1297de7aa8bc3 |
hex | 8abc2085a471 |
152540604114033 has 64 divisors (see below), whose sum is σ = 230467163996160. Its totient is φ = 89259984937728.
The previous prime is 152540604114023. The next prime is 152540604114061. The reversal of 152540604114033 is 330411406045251.
It is not a de Polignac number, because 152540604114033 - 217 = 152540603982961 is a prime.
It is a Curzon number.
It is a junction number, because it is equal to n+sod(n) for n = 152540604113982 and 152540604114000.
It is not an unprimeable number, because it can be changed into a prime (152540604114023) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (19) of ones.
It is a polite number, since it can be written in 63 ways as a sum of consecutive naturals, for example, 3766135 + ... + 17867987.
It is an arithmetic number, because the mean of its divisors is an integer number (3601049437440).
Almost surely, 2152540604114033 is an apocalyptic number.
It is an amenable number.
152540604114033 is a deficient number, since it is larger than the sum of its proper divisors (77926559882127).
152540604114033 is a wasteful number, since it uses less digits than its factorization.
152540604114033 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 14102158.
The product of its (nonzero) digits is 172800, while the sum is 39.
The spelling of 152540604114033 in words is "one hundred fifty-two trillion, five hundred forty billion, six hundred four million, one hundred fourteen thousand, thirty-three".
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