Base | Representation |
---|---|
bin | 1100010001001111101010… |
… | …0100110101110111110111 |
3 | 1202202200002021000022010011 |
4 | 3010103322210311313313 |
5 | 3232011320211140122 |
6 | 44405223233441051 |
7 | 2561435501544163 |
oct | 304237244656767 |
9 | 52680067008104 |
10 | 13490401271287 |
11 | 4331279907246 |
12 | 161a644110187 |
13 | 76b1a72b3b55 |
14 | 348d204dbca3 |
15 | 185db2772377 |
hex | c44fa935df7 |
13490401271287 has 2 divisors, whose sum is σ = 13490401271288. Its totient is φ = 13490401271286.
The previous prime is 13490401271189. The next prime is 13490401271293. The reversal of 13490401271287 is 78217210409431.
It is a strong prime.
It is an emirp because it is prime and its reverse (78217210409431) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 13490401271287 - 215 = 13490401238519 is a prime.
It is a self number, because there is not a number n which added to its sum of digits gives 13490401271287.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (13490401671287) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6745200635643 + 6745200635644.
It is an arithmetic number, because the mean of its divisors is an integer number (6745200635644).
Almost surely, 213490401271287 is an apocalyptic number.
13490401271287 is a deficient number, since it is larger than the sum of its proper divisors (1).
13490401271287 is an equidigital number, since it uses as much as digits as its factorization.
13490401271287 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 677376, while the sum is 49.
The spelling of 13490401271287 in words is "thirteen trillion, four hundred ninety billion, four hundred one million, two hundred seventy-one thousand, two hundred eighty-seven".
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