Base | Representation |
---|---|
bin | 10101010011101111111… |
… | …100111001101001110011 |
3 | 12011222122121020222210222 |
4 | 111103233330321221303 |
5 | 142442404014200201 |
6 | 3040410303443255 |
7 | 210536036506442 |
oct | 25235774715163 |
9 | 5158577228728 |
10 | 1464314600051 |
11 | 515015465216 |
12 | 1b7963bb652b |
13 | a81131c5b34 |
14 | 50c32106d59 |
15 | 2815457c91b |
hex | 154eff39a73 |
1464314600051 has 2 divisors, whose sum is σ = 1464314600052. Its totient is φ = 1464314600050.
The previous prime is 1464314600011. The next prime is 1464314600107. The reversal of 1464314600051 is 1500064134641.
It is a weak prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-1464314600051 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 1464314599984 and 1464314600020.
It is not a weakly prime, because it can be changed into another prime (1464314600011) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 732157300025 + 732157300026.
It is an arithmetic number, because the mean of its divisors is an integer number (732157300026).
Almost surely, 21464314600051 is an apocalyptic number.
1464314600051 is a deficient number, since it is larger than the sum of its proper divisors (1).
1464314600051 is an equidigital number, since it uses as much as digits as its factorization.
1464314600051 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 34560, while the sum is 35.
Adding to 1464314600051 its reverse (1500064134641), we get a palindrome (2964378734692).
The spelling of 1464314600051 in words is "one trillion, four hundred sixty-four billion, three hundred fourteen million, six hundred thousand, fifty-one".
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