Base | Representation |
---|---|
bin | 110010000001101010… |
… | …111110101110101011 |
3 | 12010122111111121022101 |
4 | 302001222332232223 |
5 | 1340002033313103 |
6 | 40402033434231 |
7 | 3611053055206 |
oct | 620152765653 |
9 | 163574447271 |
10 | 53715135403 |
11 | 20864880296 |
12 | a4b110b977 |
13 | 50b06627b7 |
14 | 2857cd093d |
15 | 15e5aeaa1d |
hex | c81abebab |
53715135403 has 2 divisors, whose sum is σ = 53715135404. Its totient is φ = 53715135402.
The previous prime is 53715135389. The next prime is 53715135431. The reversal of 53715135403 is 30453151735.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 53715135403 - 25 = 53715135371 is a prime.
It is not a weakly prime, because it can be changed into another prime (53715135493) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 26857567701 + 26857567702.
It is an arithmetic number, because the mean of its divisors is an integer number (26857567702).
Almost surely, 253715135403 is an apocalyptic number.
53715135403 is a deficient number, since it is larger than the sum of its proper divisors (1).
53715135403 is an equidigital number, since it uses as much as digits as its factorization.
53715135403 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 94500, while the sum is 37.
The spelling of 53715135403 in words is "fifty-three billion, seven hundred fifteen million, one hundred thirty-five thousand, four hundred three".
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