Base | Representation |
---|---|
bin | 100101001100101101001… |
… | …0000101111110110101111 |
3 | 200002202022200112211111101 |
4 | 1022121122100233312233 |
5 | 1132230423002000444 |
6 | 14512354534503531 |
7 | 1035240101032012 |
oct | 112313220576657 |
9 | 20082280484441 |
10 | 5112525421999 |
11 | 16a1234354596 |
12 | 6a6a12504ba7 |
13 | 2b11558518c8 |
14 | 13963a117379 |
15 | 8cec63b73d4 |
hex | 4a65a42fdaf |
5112525421999 has 2 divisors, whose sum is σ = 5112525422000. Its totient is φ = 5112525421998.
The previous prime is 5112525421867. The next prime is 5112525422063. The reversal of 5112525421999 is 9991245252115.
5112525421999 is digitally balanced in base 3, because in such base it contains all the possibile digits an equal number of times.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 5112525421999 - 217 = 5112525290927 is a prime.
It is a super-2 number, since 2×51125254219992 (a number of 26 digits) contains 22 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (5112525421799) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2556262710999 + 2556262711000.
It is an arithmetic number, because the mean of its divisors is an integer number (2556262711000).
Almost surely, 25112525421999 is an apocalyptic number.
5112525421999 is a deficient number, since it is larger than the sum of its proper divisors (1).
5112525421999 is an equidigital number, since it uses as much as digits as its factorization.
5112525421999 is an evil number, because the sum of its binary digits is even.
The product of its digits is 2916000, while the sum is 55.
The spelling of 5112525421999 in words is "five trillion, one hundred twelve billion, five hundred twenty-five million, four hundred twenty-one thousand, nine hundred ninety-nine".
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