Base | Representation |
---|---|
bin | 11001101111101110101100… |
… | …001101110111111010000111 |
3 | 112211220210021220002100002121 |
4 | 121233232230031313322013 |
5 | 104320134304202024341 |
6 | 1040453353234350411 |
7 | 32564451341544451 |
oct | 3157565415677207 |
9 | 484823256070077 |
10 | 113231112142471 |
11 | 3309605496a204 |
12 | 10848b4642b407 |
13 | 4b24856a2808b |
14 | 1dd65b4d432d1 |
15 | d156039884d1 |
hex | 66fbac377e87 |
113231112142471 has 2 divisors, whose sum is σ = 113231112142472. Its totient is φ = 113231112142470.
The previous prime is 113231112142457. The next prime is 113231112142507. The reversal of 113231112142471 is 174241211132311.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 113231112142471 - 241 = 111032088886919 is a prime.
It is a super-3 number, since 3×1132311121424713 (a number of 43 digits) contains 333 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (113231112182471) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 56615556071235 + 56615556071236.
It is an arithmetic number, because the mean of its divisors is an integer number (56615556071236).
Almost surely, 2113231112142471 is an apocalyptic number.
113231112142471 is a deficient number, since it is larger than the sum of its proper divisors (1).
113231112142471 is an equidigital number, since it uses as much as digits as its factorization.
113231112142471 is an evil number, because the sum of its binary digits is even.
The product of its digits is 8064, while the sum is 34.
Adding to 113231112142471 its reverse (174241211132311), we get a palindrome (287472323274782).
The spelling of 113231112142471 in words is "one hundred thirteen trillion, two hundred thirty-one billion, one hundred twelve million, one hundred forty-two thousand, four hundred seventy-one".
• e-mail: info -at- numbersaplenty.com • Privacy notice • done in 0.074 sec. • engine limits •