Base | Representation |
---|---|
bin | 11000000111101010110010… |
… | …11011001000010001100111 |
3 | 20221210120122101020121020111 |
4 | 30003311121123020101213 |
5 | 23423002011131003221 |
6 | 304450132505513451 |
7 | 14113005630226252 |
oct | 1403653133102147 |
9 | 227716571217214 |
10 | 53040051422311 |
11 | 1599a183305188 |
12 | 5b47625aa2887 |
13 | 2379869b6b025 |
14 | d1521b856099 |
15 | 61ea60646be1 |
hex | 303d596c8467 |
53040051422311 has 2 divisors, whose sum is σ = 53040051422312. Its totient is φ = 53040051422310.
The previous prime is 53040051422299. The next prime is 53040051422339. The reversal of 53040051422311 is 11322415004035.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 53040051422311 - 215 = 53040051389543 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (53040051822311) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 26520025711155 + 26520025711156.
It is an arithmetic number, because the mean of its divisors is an integer number (26520025711156).
Almost surely, 253040051422311 is an apocalyptic number.
53040051422311 is a deficient number, since it is larger than the sum of its proper divisors (1).
53040051422311 is an equidigital number, since it uses as much as digits as its factorization.
53040051422311 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 14400, while the sum is 31.
Adding to 53040051422311 its reverse (11322415004035), we get a palindrome (64362466426346).
The spelling of 53040051422311 in words is "fifty-three trillion, forty billion, fifty-one million, four hundred twenty-two thousand, three hundred eleven".
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