Base | Representation |
---|---|
bin | 1100010001001111101010… |
… | …0100110101110111111101 |
3 | 1202202200002021000022010101 |
4 | 3010103322210311313331 |
5 | 3232011320211140133 |
6 | 44405223233441101 |
7 | 2561435501544202 |
oct | 304237244656775 |
9 | 52680067008111 |
10 | 13490401271293 |
11 | 4331279907251 |
12 | 161a644110191 |
13 | 76b1a72b3b5b |
14 | 348d204dbca9 |
15 | 185db277237d |
hex | c44fa935dfd |
13490401271293 has 2 divisors, whose sum is σ = 13490401271294. Its totient is φ = 13490401271292.
The previous prime is 13490401271287. The next prime is 13490401271351. The reversal of 13490401271293 is 39217210409431.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 9723688868089 + 3766712403204 = 3118283^2 + 1940802^2 .
It is a cyclic number.
It is not a de Polignac number, because 13490401271293 - 229 = 13489864400381 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (13490401275293) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6745200635646 + 6745200635647.
It is an arithmetic number, because the mean of its divisors is an integer number (6745200635647).
Almost surely, 213490401271293 is an apocalyptic number.
It is an amenable number.
13490401271293 is a deficient number, since it is larger than the sum of its proper divisors (1).
13490401271293 is an equidigital number, since it uses as much as digits as its factorization.
13490401271293 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 326592, while the sum is 46.
The spelling of 13490401271293 in words is "thirteen trillion, four hundred ninety billion, four hundred one million, two hundred seventy-one thousand, two hundred ninety-three".
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