Base | Representation |
---|---|
bin | 110011000111111101000… |
… | …101000011001001111001 |
3 | 110102212021100001001012011 |
4 | 303013331011003021321 |
5 | 430030100431140441 |
6 | 11245542454245521 |
7 | 511552135214044 |
oct | 63077505031171 |
9 | 13385240031164 |
10 | 3513234240121 |
11 | 1134a55158701 |
12 | 488a7b0568a1 |
13 | 1c63b1734898 |
14 | c2081a1bc5b |
15 | 615c2148581 |
hex | 331fd143279 |
3513234240121 has 2 divisors, whose sum is σ = 3513234240122. Its totient is φ = 3513234240120.
The previous prime is 3513234240119. The next prime is 3513234240133. The reversal of 3513234240121 is 1210424323153.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 2494014770025 + 1019219470096 = 1579245^2 + 1009564^2 .
It is a cyclic number.
It is not a de Polignac number, because 3513234240121 - 21 = 3513234240119 is a prime.
Together with 3513234240119, it forms a pair of twin primes.
It is not a weakly prime, because it can be changed into another prime (3513234243121) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1756617120060 + 1756617120061.
It is an arithmetic number, because the mean of its divisors is an integer number (1756617120061).
Almost surely, 23513234240121 is an apocalyptic number.
It is an amenable number.
3513234240121 is a deficient number, since it is larger than the sum of its proper divisors (1).
3513234240121 is an equidigital number, since it uses as much as digits as its factorization.
3513234240121 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 17280, while the sum is 31.
Adding to 3513234240121 its reverse (1210424323153), we get a palindrome (4723658563274).
The spelling of 3513234240121 in words is "three trillion, five hundred thirteen billion, two hundred thirty-four million, two hundred forty thousand, one hundred twenty-one".
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