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4331231129989 is a prime number
BaseRepresentation
bin111111000001110001011…
…111100011000110000101
3120100001122221111111220101
4333001301133203012011
51031430330132124424
613113423510404101
7624630642105541
oct77016137430605
916301587444811
104331231129989
11141a9570700a5
1259b508660631
13255583092cca
1410d8c0200a21
15779ea3cc344
hex3f0717e3185

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".