Base | Representation |
---|---|
bin | 11000011010000000010011… |
… | …01010011001110011001011 |
3 | 21001000210211200211020201201 |
4 | 30031000021222121303023 |
5 | 24013312302300111413 |
6 | 310051405200125031 |
7 | 14206351311102052 |
oct | 1415001152316313 |
9 | 231023750736651 |
10 | 53670073441483 |
11 | 16112394311074 |
12 | 6029753006777 |
13 | 23c40b25ccb05 |
14 | d37907037399 |
15 | 631135edc9dd |
hex | 30d009a99ccb |
53670073441483 has 2 divisors, whose sum is σ = 53670073441484. Its totient is φ = 53670073441482.
The previous prime is 53670073441441. The next prime is 53670073441507. The reversal of 53670073441483 is 38414437007635.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-53670073441483 is a prime.
It is not a weakly prime, because it can be changed into another prime (53670073441583) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 26835036720741 + 26835036720742.
It is an arithmetic number, because the mean of its divisors is an integer number (26835036720742).
Almost surely, 253670073441483 is an apocalyptic number.
53670073441483 is a deficient number, since it is larger than the sum of its proper divisors (1).
53670073441483 is an equidigital number, since it uses as much as digits as its factorization.
53670073441483 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 20321280, while the sum is 55.
The spelling of 53670073441483 in words is "fifty-three trillion, six hundred seventy billion, seventy-three million, four hundred forty-one thousand, four hundred eighty-three".
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