Base | Representation |
---|---|
bin | 11110001111110110111000… |
… | …101010001000011010000001 |
3 | 122110000122120111021000100021 |
4 | 132033312320222020122001 |
5 | 114414040213222024413 |
6 | 1150533353005253441 |
7 | 40010116112432041 |
oct | 3617667052103201 |
9 | 573018514230307 |
10 | 133031120111233 |
11 | 3942a1aa2090a3 |
12 | 12b063a4b64281 |
13 | 592ca1c54ca96 |
14 | 24bca52195721 |
15 | 105a6a0e2118d |
hex | 78fdb8a88681 |
133031120111233 has 4 divisors (see below), whose sum is σ = 136124867090608. Its totient is φ = 129937373131860.
The previous prime is 133031120111203. The next prime is 133031120111237. The reversal of 133031120111233 is 332111021130331.
It is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-133031120111233 is a prime.
It is a Duffinian number.
It is a junction number, because it is equal to n+sod(n) for n = 133031120111198 and 133031120111207.
It is not an unprimeable number, because it can be changed into a prime (133031120111237) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (23) of ones.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 1546873489623 + ... + 1546873489708.
It is an arithmetic number, because the mean of its divisors is an integer number (34031216772652).
Almost surely, 2133031120111233 is an apocalyptic number.
It is an amenable number.
133031120111233 is a deficient number, since it is larger than the sum of its proper divisors (3093746979375).
133031120111233 is an equidigital number, since it uses as much as digits as its factorization.
133031120111233 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 3093746979374.
The product of its (nonzero) digits is 972, while the sum is 25.
Adding to 133031120111233 its reverse (332111021130331), we get a palindrome (465142141241564).
The spelling of 133031120111233 in words is "one hundred thirty-three trillion, thirty-one billion, one hundred twenty million, one hundred eleven thousand, two hundred thirty-three".
• e-mail: info -at- numbersaplenty.com • Privacy notice • done in 0.086 sec. • engine limits •