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399312000000 = 21032564759
BaseRepresentation
bin1011100111110001101…
…10011001010000000000
31102011200201222100102100
411303320312121100000
523020242333000000
6503235201550400
740564215666054
oct5637066312000
91364621870370
10399312000000
11144390239861
1265480945400
132b868c40c54
1415480a6dc64
15a5c12e6a00
hex5cf8d99400

399312000000 has 924 divisors, whose sum is σ = 1496849590080. Its totient is φ = 102451200000.

The previous prime is 399311999963. The next prime is 399312000053. The reversal of 399312000000 is 213993.

It is a congruent number.

It is an unprimeable number.

It is a pernicious number, because its binary representation contains a prime number (17) of ones.

It is a polite number, since it can be written in 83 ways as a sum of consecutive naturals, for example, 6767999971 + ... + 6768000029.

Almost surely, 2399312000000 is an apocalyptic number.

399312000000 is a gapful number since it is divisible by the number (30) formed by its first and last digit.

It is an amenable number.

It is a practical number, because each smaller number is the sum of distinct divisors of 399312000000, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (748424795040).

399312000000 is an abundant number, since it is smaller than the sum of its proper divisors (1097537590080).

It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.

399312000000 is an frugal number, since it uses more digits than its factorization.

399312000000 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 162 (or 116 counting only the distinct ones).

The product of its (nonzero) digits is 1458, while the sum is 27.

Adding to 399312000000 its reverse (213993), we get a palindrome (399312213993).

The spelling of 399312000000 in words is "three hundred ninety-nine billion, three hundred twelve million", and thus it is an aban number.