Base | Representation |
---|---|
bin | 100111000001000001011… |
… | …0001001111001010010011 |
3 | 200222122001211012011121211 |
4 | 1032002002301033022103 |
5 | 1200324004130140033 |
6 | 15223225044200551 |
7 | 1062262055362360 |
oct | 116020261171223 |
9 | 20878054164554 |
10 | 5362313130643 |
11 | 1788164a70137 |
12 | 727303999157 |
13 | 2cb8829419c1 |
14 | 147774667467 |
15 | 9474574a2cd |
hex | 4e082c4f293 |
5362313130643 has 16 divisors (see below), whose sum is σ = 6141532320000. Its totient is φ = 4586397459552.
The previous prime is 5362313130623. The next prime is 5362313130667. The reversal of 5362313130643 is 3460313132635.
It is a cyclic number.
It is not a de Polignac number, because 5362313130643 - 233 = 5353723196051 is a prime.
It is a super-2 number, since 2×53623131306432 (a number of 26 digits) contains 22 as substring.
It is a junction number, because it is equal to n+sod(n) for n = 5362313130596 and 5362313130605.
It is not an unprimeable number, because it can be changed into a prime (5362313130623) by changing a digit.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 6101133 + ... + 6924481.
It is an arithmetic number, because the mean of its divisors is an integer number (383845770000).
Almost surely, 25362313130643 is an apocalyptic number.
5362313130643 is a deficient number, since it is larger than the sum of its proper divisors (779219189357).
5362313130643 is a wasteful number, since it uses less digits than its factorization.
5362313130643 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 825354.
The product of its (nonzero) digits is 349920, while the sum is 40.
The spelling of 5362313130643 in words is "five trillion, three hundred sixty-two billion, three hundred thirteen million, one hundred thirty thousand, six hundred forty-three".
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