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30421311300 = 2232521719710093
BaseRepresentation
bin11100010101010000…
…000101111101000100
32220112010001201211100
4130111100011331010
5444300323430200
621550402140100
72124603610452
oct342520057504
986463051740
1030421311300
11119a1045496
125a90051630
132b3a758923
1416883999d2
15bd0aebd00
hex715405f44

30421311300 has 216 divisors, whose sum is σ = 101485520136. Its totient is φ = 7595642880.

The previous prime is 30421311299. The next prime is 30421311329. The reversal of 30421311300 is 311312403.

30421311300 is a `hidden beast` number, since 30 + 4 + 21 + 311 + 300 = 666.

It can be written as a sum of positive squares in 12 ways, for example, as 111894084 + 30309417216 = 10578^2 + 174096^2 .

It is a Harshad number since it is a multiple of its sum of digits (18).

It is an unprimeable number.

It is a polite number, since it can be written in 71 ways as a sum of consecutive naturals, for example, 3009054 + ... + 3019146.

It is an arithmetic number, because the mean of its divisors is an integer number (469840371).

Almost surely, 230421311300 is an apocalyptic number.

30421311300 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 30421311300, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (50742760068).

30421311300 is an abundant number, since it is smaller than the sum of its proper divisors (71064208836).

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

30421311300 is a wasteful number, since it uses less digits than its factorization.

30421311300 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 10327 (or 10317 counting only the distinct ones).

The product of its (nonzero) digits is 216, while the sum is 18.

Adding to 30421311300 its reverse (311312403), we get a palindrome (30732623703).

It can be divided in two parts, 304213 and 11300, that added together give a palindrome (315513).

The spelling of 30421311300 in words is "thirty billion, four hundred twenty-one million, three hundred eleven thousand, three hundred".