Base | Representation |
---|---|
bin | 1011110101010110100010… |
… | …0010010000111001011001 |
3 | 1201001212012120121221101011 |
4 | 2331111220202100321121 |
5 | 3201133424144032103 |
6 | 43401133420253521 |
7 | 2512012623046504 |
oct | 275255042207131 |
9 | 51055176557334 |
10 | 13011209752153 |
11 | 4167028781478 |
12 | 15617ab9a12a1 |
13 | 734c4b1a2b62 |
14 | 32da62ad613b |
15 | 1786b8859d6d |
hex | bd568890e59 |
13011209752153 has 2 divisors, whose sum is σ = 13011209752154. Its totient is φ = 13011209752152.
The previous prime is 13011209752069. The next prime is 13011209752271. The reversal of 13011209752153 is 35125790211031.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 11487239254089 + 1523970498064 = 3389283^2 + 1234492^2 .
It is a cyclic number.
It is not a de Polignac number, because 13011209752153 - 221 = 13011207655001 is a prime.
It is not a weakly prime, because it can be changed into another prime (13011209759153) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6505604876076 + 6505604876077.
It is an arithmetic number, because the mean of its divisors is an integer number (6505604876077).
Almost surely, 213011209752153 is an apocalyptic number.
It is an amenable number.
13011209752153 is a deficient number, since it is larger than the sum of its proper divisors (1).
13011209752153 is an equidigital number, since it uses as much as digits as its factorization.
13011209752153 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 56700, while the sum is 40.
Adding to 13011209752153 its reverse (35125790211031), we get a palindrome (48136999963184).
The spelling of 13011209752153 in words is "thirteen trillion, eleven billion, two hundred nine million, seven hundred fifty-two thousand, one hundred fifty-three".
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