Base | Representation |
---|---|
bin | 1000111011001100110… |
… | …0000111111001111011 |
3 | 112122202211201220220111 |
4 | 2032303030013321323 |
5 | 10003010004431011 |
6 | 154234445205151 |
7 | 14035443546226 |
oct | 2166314077173 |
9 | 478684656814 |
10 | 153330155131 |
11 | 5a032969a7a |
12 | 25871b607b7 |
13 | 115c748c25a |
14 | 75c7b7a5bd |
15 | 3ec613c121 |
hex | 23b3307e7b |
153330155131 has 2 divisors, whose sum is σ = 153330155132. Its totient is φ = 153330155130.
The previous prime is 153330155119. The next prime is 153330155159. The reversal of 153330155131 is 131551033351.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 153330155131 - 27 = 153330155003 is a prime.
It is a super-3 number, since 3×1533301551313 (a number of 35 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.
It is a junction number, because it is equal to n+sod(n) for n = 153330155093 and 153330155102.
It is not a weakly prime, because it can be changed into another prime (153330155171) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 76665077565 + 76665077566.
It is an arithmetic number, because the mean of its divisors is an integer number (76665077566).
Almost surely, 2153330155131 is an apocalyptic number.
153330155131 is a deficient number, since it is larger than the sum of its proper divisors (1).
153330155131 is an equidigital number, since it uses as much as digits as its factorization.
153330155131 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 10125, while the sum is 31.
Adding to 153330155131 its reverse (131551033351), we get a palindrome (284881188482).
The spelling of 153330155131 in words is "one hundred fifty-three billion, three hundred thirty million, one hundred fifty-five thousand, one hundred thirty-one".
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