Base | Representation |
---|---|
bin | 111111000001110001011… |
… | …111100011000110000101 |
3 | 120100001122221111111220101 |
4 | 333001301133203012011 |
5 | 1031430330132124424 |
6 | 13113423510404101 |
7 | 624630642105541 |
oct | 77016137430605 |
9 | 16301587444811 |
10 | 4331231129989 |
11 | 141a9570700a5 |
12 | 59b508660631 |
13 | 255583092cca |
14 | 10d8c0200a21 |
15 | 779ea3cc344 |
hex | 3f0717e3185 |
4331231129989 has 2 divisors, whose sum is σ = 4331231129990. Its totient is φ = 4331231129988.
The previous prime is 4331231129959. The next prime is 4331231130041. The reversal of 4331231129989 is 9899211321334.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 2490582672964 + 1840648457025 = 1578158^2 + 1356705^2 .
It is a cyclic number.
It is not a de Polignac number, because 4331231129989 - 25 = 4331231129957 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (4331231129959) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2165615564994 + 2165615564995.
It is an arithmetic number, because the mean of its divisors is an integer number (2165615564995).
Almost surely, 24331231129989 is an apocalyptic number.
It is an amenable number.
4331231129989 is a deficient number, since it is larger than the sum of its proper divisors (1).
4331231129989 is an equidigital number, since it uses as much as digits as its factorization.
4331231129989 is an evil number, because the sum of its binary digits is even.
The product of its digits is 2519424, while the sum is 55.
The spelling of 4331231129989 in words is "four trillion, three hundred thirty-one billion, two hundred thirty-one million, one hundred twenty-nine thousand, nine hundred eighty-nine".
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