Base | Representation |
---|---|
bin | 10110010101101000011… |
… | …011101101101011100011 |
3 | 12102202020122120222210022 |
4 | 112111220123231223203 |
5 | 200122243141421011 |
6 | 3133110021024055 |
7 | 215622050532422 |
oct | 26255033555343 |
9 | 5382218528708 |
10 | 1535055420131 |
11 | 542016847392 |
12 | 209606b5502b |
13 | b19a7c7275c |
14 | 544232812b9 |
15 | 29de4c258db |
hex | 165686edae3 |
1535055420131 has 2 divisors, whose sum is σ = 1535055420132. Its totient is φ = 1535055420130.
The previous prime is 1535055420103. The next prime is 1535055420181. The reversal of 1535055420131 is 1310245505351.
It is a weak prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-1535055420131 is a prime.
It is a super-2 number, since 2×15350554201312 (a number of 25 digits) contains 22 as substring.
It is a junction number, because it is equal to n+sod(n) for n = 1535055420091 and 1535055420100.
It is not a weakly prime, because it can be changed into another prime (1535055420181) 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 as a sum of consecutive naturals, namely, 767527710065 + 767527710066.
It is an arithmetic number, because the mean of its divisors is an integer number (767527710066).
Almost surely, 21535055420131 is an apocalyptic number.
1535055420131 is a deficient number, since it is larger than the sum of its proper divisors (1).
1535055420131 is an equidigital number, since it uses as much as digits as its factorization.
1535055420131 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 45000, while the sum is 35.
The spelling of 1535055420131 in words is "one trillion, five hundred thirty-five billion, fifty-five million, four hundred twenty thousand, one hundred thirty-one".
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