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300011302113 = 3209947643529
BaseRepresentation
bin1000101110110100001…
…00010010110011100001
31001200101022012020022220
410113122010102303201
514403410343131423
6345453455221253
730450362664165
oct4273204226341
91050338166286
10300011302113
11106263616796
124a189217829
13223a2288053
1410740844ca5
157c0d6929e3
hex45da112ce1

300011302113 has 8 divisors (see below), whose sum is σ = 400205652000. Its totient is φ = 199912243488.

The previous prime is 300011302093. The next prime is 300011302147. The reversal of 300011302113 is 311203110003.

It is a sphenic number, since it is the product of 3 distinct primes.

It is not a de Polignac number, because 300011302113 - 210 = 300011301089 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 300011302092 and 300011302101.

It is not an unprimeable number, because it can be changed into a prime (300011304113) by changing a digit.

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 7 ways as a sum of consecutive naturals, for example, 23815468 + ... + 23828061.

It is an arithmetic number, because the mean of its divisors is an integer number (50025706500).

Almost surely, 2300011302113 is an apocalyptic number.

It is an amenable number.

300011302113 is a deficient number, since it is larger than the sum of its proper divisors (100194349887).

300011302113 is a wasteful number, since it uses less digits than its factorization.

300011302113 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 47645631.

The product of its (nonzero) digits is 54, while the sum is 15.

Adding to 300011302113 its reverse (311203110003), we get a palindrome (611214412116).

The spelling of 300011302113 in words is "three hundred billion, eleven million, three hundred two thousand, one hundred thirteen".

Divisors: 1 3 2099 6297 47643529 142930587 100003767371 300011302113