Base | Representation |
---|---|
bin | 1011111011001110011101… |
… | …1000000101100000111011 |
3 | 1201102111200022220012222211 |
4 | 2332303213120011200323 |
5 | 3204312111220211311 |
6 | 43515343052555551 |
7 | 2522214415244461 |
oct | 276634730054073 |
9 | 51374608805884 |
10 | 13112122038331 |
11 | 41a5902070698 |
12 | 1579273182bb7 |
13 | 7416109bb2b4 |
14 | 3348b6d90831 |
15 | 17b122beb821 |
hex | bece760583b |
13112122038331 has 2 divisors, whose sum is σ = 13112122038332. Its totient is φ = 13112122038330.
The previous prime is 13112122038293. The next prime is 13112122038343. The reversal of 13112122038331 is 13383022121131.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 13112122038331 - 27 = 13112122038203 is a prime.
It is a super-2 number, since 2×131121220383312 (a number of 27 digits) contains 22 as substring.
It is a junction number, because it is equal to n+sod(n) for n = 13112122038293 and 13112122038302.
It is not a weakly prime, because it can be changed into another prime (13112122038391) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6556061019165 + 6556061019166.
It is an arithmetic number, because the mean of its divisors is an integer number (6556061019166).
Almost surely, 213112122038331 is an apocalyptic number.
13112122038331 is a deficient number, since it is larger than the sum of its proper divisors (1).
13112122038331 is an equidigital number, since it uses as much as digits as its factorization.
13112122038331 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 5184, while the sum is 31.
Adding to 13112122038331 its reverse (13383022121131), we get a palindrome (26495144159462).
The spelling of 13112122038331 in words is "thirteen trillion, one hundred twelve billion, one hundred twenty-two million, thirty-eight thousand, three hundred thirty-one".
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