Base | Representation |
---|---|
bin | 1100100110110001010… |
… | …01001111111000101111 |
3 | 1112101222200011221202001 |
4 | 12103120221033320233 |
5 | 24044023120434411 |
6 | 530551113313131 |
7 | 43202254222555 |
oct | 6233051177057 |
9 | 1471880157661 |
10 | 433131421231 |
11 | 157765439563 |
12 | 6bb3aa057a7 |
13 | 31ac86b02c2 |
14 | 16d6c463bd5 |
15 | b4003c5dc1 |
hex | 64d8a4fe2f |
433131421231 has 2 divisors, whose sum is σ = 433131421232. Its totient is φ = 433131421230.
The previous prime is 433131421223. The next prime is 433131421397. The reversal of 433131421231 is 132124131334.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 433131421231 - 23 = 433131421223 is a prime.
It is a super-2 number, since 2×4331314212312 (a number of 24 digits) contains 22 as substring.
It is a self number, because there is not a number n which added to its sum of digits gives 433131421231.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (433131491231) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 216565710615 + 216565710616.
It is an arithmetic number, because the mean of its divisors is an integer number (216565710616).
Almost surely, 2433131421231 is an apocalyptic number.
433131421231 is a deficient number, since it is larger than the sum of its proper divisors (1).
433131421231 is an equidigital number, since it uses as much as digits as its factorization.
433131421231 is an evil number, because the sum of its binary digits is even.
The product of its digits is 5184, while the sum is 28.
Adding to 433131421231 its reverse (132124131334), we get a palindrome (565255552565).
The spelling of 433131421231 in words is "four hundred thirty-three billion, one hundred thirty-one million, four hundred twenty-one thousand, two hundred thirty-one".
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