Base | Representation |
---|---|
bin | 1100001100010011100000… |
… | …1010101001001101011011 |
3 | 1202110120000210121111010211 |
4 | 3003010320022221031123 |
5 | 3224114014010321042 |
6 | 44302230132151551 |
7 | 2552342442326353 |
oct | 303047012511533 |
9 | 52416023544124 |
10 | 13405535245147 |
11 | 42a928a189761 |
12 | 16060ba7685b7 |
13 | 7631a0cca936 |
14 | 344b8d3b7b63 |
15 | 183a96eb3017 |
hex | c31382a935b |
13405535245147 has 2 divisors, whose sum is σ = 13405535245148. Its totient is φ = 13405535245146.
The previous prime is 13405535245129. The next prime is 13405535245177. The reversal of 13405535245147 is 74154253550431.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 13405535245147 - 27 = 13405535245019 is a prime.
It is a self number, because there is not a number n which added to its sum of digits gives 13405535245147.
It is not a weakly prime, because it can be changed into another prime (13405535245177) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6702767622573 + 6702767622574.
It is an arithmetic number, because the mean of its divisors is an integer number (6702767622574).
Almost surely, 213405535245147 is an apocalyptic number.
13405535245147 is a deficient number, since it is larger than the sum of its proper divisors (1).
13405535245147 is an equidigital number, since it uses as much as digits as its factorization.
13405535245147 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 5040000, while the sum is 49.
Adding to 13405535245147 its reverse (74154253550431), we get a palindrome (87559788795578).
The spelling of 13405535245147 in words is "thirteen trillion, four hundred five billion, five hundred thirty-five million, two hundred forty-five thousand, one hundred forty-seven".
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