Base | Representation |
---|---|
bin | 1011000001001100000000… |
… | …1001101111010010101001 |
3 | 1120220012000101020110120111 |
4 | 2300103000021233102221 |
5 | 3041443041034200233 |
6 | 41433322455254321 |
7 | 2360165530045162 |
oct | 260230011572251 |
9 | 46805011213514 |
10 | 12115031553193 |
11 | 3950a57523338 |
12 | 1437b8367a3a1 |
13 | 69b59a4452aa |
14 | 2dc529118769 |
15 | 160216b03acd |
hex | b04c026f4a9 |
12115031553193 has 2 divisors, whose sum is σ = 12115031553194. Its totient is φ = 12115031553192.
The previous prime is 12115031553137. The next prime is 12115031553203. The reversal of 12115031553193 is 39135513051121.
It is a happy number.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 7895470572544 + 4219560980649 = 2809888^2 + 2054157^2 .
It is a cyclic number.
It is not a de Polignac number, because 12115031553193 - 229 = 12114494682281 is a prime.
It is not a weakly prime, because it can be changed into another prime (12115031553133) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6057515776596 + 6057515776597.
It is an arithmetic number, because the mean of its divisors is an integer number (6057515776597).
Almost surely, 212115031553193 is an apocalyptic number.
It is an amenable number.
12115031553193 is a deficient number, since it is larger than the sum of its proper divisors (1).
12115031553193 is an equidigital number, since it uses as much as digits as its factorization.
12115031553193 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 60750, while the sum is 40.
The spelling of 12115031553193 in words is "twelve trillion, one hundred fifteen billion, thirty-one million, five hundred fifty-three thousand, one hundred ninety-three".
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