Base | Representation |
---|---|
bin | 11110000010100000110110… |
… | …000010000001001110111011 |
3 | 122022202222120211010022111212 |
4 | 132002200312002001032323 |
5 | 114304024134213231323 |
6 | 1144552214021113335 |
7 | 36553633366520405 |
oct | 3602406602011673 |
9 | 568688524108455 |
10 | 132114100523963 |
11 | 391063035a14a8 |
12 | 12998720b5884b |
13 | 58943c9a02c9c |
14 | 248a4dc671b75 |
15 | 10418ce64db78 |
hex | 7828360813bb |
132114100523963 has 2 divisors, whose sum is σ = 132114100523964. Its totient is φ = 132114100523962.
The previous prime is 132114100523929. The next prime is 132114100523969. The reversal of 132114100523963 is 369325001411231.
It is a strong prime.
It is an emirp because it is prime and its reverse (369325001411231) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 132114100523963 - 210 = 132114100522939 is a prime.
It is a super-2 number, since 2×1321141005239632 (a number of 29 digits) contains 22 as substring.
It is a self number, because there is not a number n which added to its sum of digits gives 132114100523963.
It is not a weakly prime, because it can be changed into another prime (132114100523969) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 66057050261981 + 66057050261982.
It is an arithmetic number, because the mean of its divisors is an integer number (66057050261982).
Almost surely, 2132114100523963 is an apocalyptic number.
132114100523963 is a deficient number, since it is larger than the sum of its proper divisors (1).
132114100523963 is an equidigital number, since it uses as much as digits as its factorization.
132114100523963 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 116640, while the sum is 41.
The spelling of 132114100523963 in words is "one hundred thirty-two trillion, one hundred fourteen billion, one hundred million, five hundred twenty-three thousand, nine hundred sixty-three".
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