Base | Representation |
---|---|
bin | 11000011010000000010011… |
… | …01010011001110010100001 |
3 | 21001000210211200211020200011 |
4 | 30031000021222121302201 |
5 | 24013312302300111231 |
6 | 310051405200124521 |
7 | 14206351311101662 |
oct | 1415001152316241 |
9 | 231023750736604 |
10 | 53670073441441 |
11 | 16112394311036 |
12 | 6029753006741 |
13 | 23c40b25ccaa2 |
14 | d37907037369 |
15 | 631135edc9b1 |
hex | 30d009a99ca1 |
53670073441441 has 2 divisors, whose sum is σ = 53670073441442. Its totient is φ = 53670073441440.
The previous prime is 53670073441409. The next prime is 53670073441483. The reversal of 53670073441441 is 14414437007635.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 46830823323025 + 6839250118416 = 6843305^2 + 2615196^2 .
It is a cyclic number.
It is not a de Polignac number, because 53670073441441 - 25 = 53670073441409 is a prime.
It is not a weakly prime, because it can be changed into another prime (53670073441141) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 26835036720720 + 26835036720721.
It is an arithmetic number, because the mean of its divisors is an integer number (26835036720721).
Almost surely, 253670073441441 is an apocalyptic number.
It is an amenable number.
53670073441441 is a deficient number, since it is larger than the sum of its proper divisors (1).
53670073441441 is an equidigital number, since it uses as much as digits as its factorization.
53670073441441 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 3386880, while the sum is 49.
The spelling of 53670073441441 in words is "fifty-three trillion, six hundred seventy billion, seventy-three million, four hundred forty-one thousand, four hundred forty-one".
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