Base | Representation |
---|---|
bin | 101110000100010110011… |
… | …0101011100101111001011 |
3 | 211102021210111020220201211 |
4 | 1130020230311130233023 |
5 | 1312213440201100042 |
6 | 21244355551153551 |
7 | 1222303213021003 |
oct | 134105465345713 |
9 | 24367714226654 |
10 | 6331534003147 |
11 | 2021208977a81 |
12 | 863115b158b7 |
13 | 36c0a42bcb62 |
14 | 17c63adb1003 |
15 | aea6ec85017 |
hex | 5c22cd5cbcb |
6331534003147 has 4 divisors (see below), whose sum is σ = 6331708035360. Its totient is φ = 6331359970936.
The previous prime is 6331534003027. The next prime is 6331534003219. The reversal of 6331534003147 is 7413004351336.
It is a semiprime because it is the product of two primes.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-6331534003147 is a prime.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (6331534033147) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (23) of ones.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 86961523 + ... + 87034300.
It is an arithmetic number, because the mean of its divisors is an integer number (1582927008840).
Almost surely, 26331534003147 is an apocalyptic number.
6331534003147 is a deficient number, since it is larger than the sum of its proper divisors (174032213).
6331534003147 is a wasteful number, since it uses less digits than its factorization.
6331534003147 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 174032212.
The product of its (nonzero) digits is 272160, while the sum is 40.
The spelling of 6331534003147 in words is "six trillion, three hundred thirty-one billion, five hundred thirty-four million, three thousand, one hundred forty-seven".
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