Base | Representation |
---|---|
bin | 10110010011110011111… |
… | …101000101101110111011 |
3 | 12102120012110110002221002 |
4 | 112103303331011232323 |
5 | 200104242404011011 |
6 | 3132144052035215 |
7 | 215522451325154 |
oct | 26236375055673 |
9 | 5376173402832 |
10 | 1533101235131 |
11 | 541203752264 |
12 | 209160602b0b |
13 | b175614997b |
14 | 542b991092b |
15 | 29d2d3b2c3b |
hex | 164f3f45bbb |
1533101235131 has 2 divisors, whose sum is σ = 1533101235132. Its totient is φ = 1533101235130.
The previous prime is 1533101235121. The next prime is 1533101235137. The reversal of 1533101235131 is 1315321013351.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-1533101235131 is a prime.
It is a super-2 number, since 2×15331012351312 (a number of 25 digits) contains 22 as substring.
It is a junction number, because it is equal to n+sod(n) for n = 1533101235094 and 1533101235103.
It is not a weakly prime, because it can be changed into another prime (1533101235137) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 766550617565 + 766550617566.
It is an arithmetic number, because the mean of its divisors is an integer number (766550617566).
Almost surely, 21533101235131 is an apocalyptic number.
1533101235131 is a deficient number, since it is larger than the sum of its proper divisors (1).
1533101235131 is an equidigital number, since it uses as much as digits as its factorization.
1533101235131 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 4050, while the sum is 29.
Adding to 1533101235131 its reverse (1315321013351), we get a palindrome (2848422248482).
The spelling of 1533101235131 in words is "one trillion, five hundred thirty-three billion, one hundred one million, two hundred thirty-five thousand, one hundred thirty-one".
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