Base | Representation |
---|---|
bin | 1101111101001100100110… |
… | …0101110100010000111011 |
3 | 2000022222010210201022001001 |
4 | 3133103021211310100323 |
5 | 4002403032330041011 |
6 | 52345221554505431 |
7 | 3142432352035312 |
oct | 337231145642073 |
9 | 60288123638031 |
10 | 15345005315131 |
11 | 4986864503198 |
12 | 1879b6b295277 |
13 | 8740490a086b |
14 | 3b09b8c6b279 |
15 | 1b925ae617c1 |
hex | df4c997443b |
15345005315131 has 2 divisors, whose sum is σ = 15345005315132. Its totient is φ = 15345005315130.
The previous prime is 15345005315119. The next prime is 15345005315141. The reversal of 15345005315131 is 13151350054351.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 15345005315131 - 25 = 15345005315099 is a prime.
It is a super-2 number, since 2×153450053151312 (a number of 27 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (15345005315141) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 7672502657565 + 7672502657566.
It is an arithmetic number, because the mean of its divisors is an integer number (7672502657566).
Almost surely, 215345005315131 is an apocalyptic number.
15345005315131 is a deficient number, since it is larger than the sum of its proper divisors (1).
15345005315131 is an equidigital number, since it uses as much as digits as its factorization.
15345005315131 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 67500, while the sum is 37.
Adding to 15345005315131 its reverse (13151350054351), we get a palindrome (28496355369482).
The spelling of 15345005315131 in words is "fifteen trillion, three hundred forty-five billion, five million, three hundred fifteen thousand, one hundred thirty-one".
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