Base | Representation |
---|---|
bin | 110010000001101010… |
… | …111101101100001011 |
3 | 12010122111111101110202 |
4 | 302001222331230023 |
5 | 1340002033144042 |
6 | 40402033402415 |
7 | 3611053036616 |
oct | 620152755413 |
9 | 163574441422 |
10 | 53715131147 |
11 | 20864878077 |
12 | a4b110940b |
13 | 50b0660892 |
14 | 2857ccd17d |
15 | 15e5ae9632 |
hex | c81abdb0b |
53715131147 has 2 divisors, whose sum is σ = 53715131148. Its totient is φ = 53715131146.
The previous prime is 53715131143. The next prime is 53715131207. The reversal of 53715131147 is 74113151735.
53715131147 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 53715131147 - 22 = 53715131143 is a prime.
It is a self number, because there is not a number n which added to its sum of digits gives 53715131147.
It is not a weakly prime, because it can be changed into another prime (53715131143) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 26857565573 + 26857565574.
It is an arithmetic number, because the mean of its divisors is an integer number (26857565574).
Almost surely, 253715131147 is an apocalyptic number.
53715131147 is a deficient number, since it is larger than the sum of its proper divisors (1).
53715131147 is an equidigital number, since it uses as much as digits as its factorization.
53715131147 is an evil number, because the sum of its binary digits is even.
The product of its digits is 44100, while the sum is 38.
The spelling of 53715131147 in words is "fifty-three billion, seven hundred fifteen million, one hundred thirty-one thousand, one hundred forty-seven".
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