Base | Representation |
---|---|
bin | 1101110111000110101100… |
… | …0111101001000000110111 |
3 | 1222221221222120120212102202 |
4 | 3131301223013221000313 |
5 | 3444144204312200234 |
6 | 52225152134055115 |
7 | 3132035601123263 |
oct | 335615307510067 |
9 | 58857876525382 |
10 | 15240341131319 |
11 | 4946435287827 |
12 | 186181b4aa49b |
13 | 867209190755 |
14 | 3a98ca5a30a3 |
15 | 1b668250677e |
hex | ddc6b1e9037 |
15240341131319 has 2 divisors, whose sum is σ = 15240341131320. Its totient is φ = 15240341131318.
The previous prime is 15240341131307. The next prime is 15240341131321. The reversal of 15240341131319 is 91313114304251.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-15240341131319 is a prime.
It is a super-2 number, since 2×152403411313192 (a number of 27 digits) contains 22 as substring.
Together with 15240341131321, it forms a pair of twin primes.
It is a Chen prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (15240341131369) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 7620170565659 + 7620170565660.
It is an arithmetic number, because the mean of its divisors is an integer number (7620170565660).
Almost surely, 215240341131319 is an apocalyptic number.
15240341131319 is a deficient number, since it is larger than the sum of its proper divisors (1).
15240341131319 is an equidigital number, since it uses as much as digits as its factorization.
15240341131319 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 38880, while the sum is 38.
The spelling of 15240341131319 in words is "fifteen trillion, two hundred forty billion, three hundred forty-one million, one hundred thirty-one thousand, three hundred nineteen".
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