Base | Representation |
---|---|
bin | 1100010001001011000000… |
… | …1000111110110100110101 |
3 | 1202202112212020010210202212 |
4 | 3010102300020332310311 |
5 | 3232001241012011231 |
6 | 44404455303143205 |
7 | 2561362526015243 |
oct | 304226010766465 |
9 | 52675766123685 |
10 | 13489152453941 |
11 | 433079899829a |
12 | 161a355a44b05 |
13 | 76b038659688 |
14 | 348c42701793 |
15 | 185d3cce242b |
hex | c44b023ed35 |
13489152453941 has 2 divisors, whose sum is σ = 13489152453942. Its totient is φ = 13489152453940.
The previous prime is 13489152453893. The next prime is 13489152453977. The reversal of 13489152453941 is 14935425198431.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 9491612075716 + 3997540378225 = 3080846^2 + 1999385^2 .
It is a cyclic number.
It is not a de Polignac number, because 13489152453941 - 234 = 13471972584757 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (13489152453041) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6744576226970 + 6744576226971.
It is an arithmetic number, because the mean of its divisors is an integer number (6744576226971).
Almost surely, 213489152453941 is an apocalyptic number.
It is an amenable number.
13489152453941 is a deficient number, since it is larger than the sum of its proper divisors (1).
13489152453941 is an equidigital number, since it uses as much as digits as its factorization.
13489152453941 is an evil number, because the sum of its binary digits is even.
The product of its digits is 18662400, while the sum is 59.
The spelling of 13489152453941 in words is "thirteen trillion, four hundred eighty-nine billion, one hundred fifty-two million, four hundred fifty-three thousand, nine hundred forty-one".
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