Base | Representation |
---|---|
bin | 11000010010100110001010… |
… | …10110111000111010110111 |
3 | 21000010110210201220101120202 |
4 | 30021103011112320322313 |
5 | 24000124443221424134 |
6 | 305334430402551115 |
7 | 14152100330016035 |
oct | 1411230526707267 |
9 | 230113721811522 |
10 | 53415524404919 |
11 | 1602443aa94376 |
12 | 5ba835306849b |
13 | 23a60a6c58711 |
14 | d2947a3d3555 |
15 | 6296d8b4ad7e |
hex | 3094c55b8eb7 |
53415524404919 has 2 divisors, whose sum is σ = 53415524404920. Its totient is φ = 53415524404918.
The previous prime is 53415524404789. The next prime is 53415524404939. The reversal of 53415524404919 is 91940442551435.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-53415524404919 is a prime.
It is a super-3 number, since 3×534155244049193 (a number of 42 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (53415524404939) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 26707762202459 + 26707762202460.
It is an arithmetic number, because the mean of its divisors is an integer number (26707762202460).
Almost surely, 253415524404919 is an apocalyptic number.
53415524404919 is a deficient number, since it is larger than the sum of its proper divisors (1).
53415524404919 is an equidigital number, since it uses as much as digits as its factorization.
53415524404919 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 15552000, while the sum is 56.
The spelling of 53415524404919 in words is "fifty-three trillion, four hundred fifteen billion, five hundred twenty-four million, four hundred four thousand, nine hundred nineteen".
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