Base | Representation |
---|---|
bin | 1001010110001001110… |
… | …01111110101000101001 |
3 | 1010200220011211210022222 |
4 | 10223010321332220221 |
5 | 20230134031104001 |
6 | 403305310123425 |
7 | 32125633264646 |
oct | 4530471765051 |
9 | 1120804753288 |
10 | 321131113001 |
11 | 114211161256 |
12 | 522a21b0575 |
13 | 2438995b97a |
14 | 11785744dcd |
15 | 85478a961b |
hex | 4ac4e7ea29 |
321131113001 has 2 divisors, whose sum is σ = 321131113002. Its totient is φ = 321131113000.
The previous prime is 321131112989. The next prime is 321131113049. The reversal of 321131113001 is 100311131123.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 182285448601 + 138845664400 = 426949^2 + 372620^2 .
It is a cyclic number.
It is not a de Polignac number, because 321131113001 - 214 = 321131096617 is a prime.
It is a super-2 number, since 2×3211311130012 (a number of 24 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (321131103001) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 160565556500 + 160565556501.
It is an arithmetic number, because the mean of its divisors is an integer number (160565556501).
Almost surely, 2321131113001 is an apocalyptic number.
It is an amenable number.
321131113001 is a deficient number, since it is larger than the sum of its proper divisors (1).
321131113001 is an equidigital number, since it uses as much as digits as its factorization.
321131113001 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 54, while the sum is 17.
Adding to 321131113001 its reverse (100311131123), we get a palindrome (421442244124).
The spelling of 321131113001 in words is "three hundred twenty-one billion, one hundred thirty-one million, one hundred thirteen thousand, one".
• e-mail: info -at- numbersaplenty.com • Privacy notice • done in 0.091 sec. • engine limits •