Base | Representation |
---|---|
bin | 1011111011111011111101… |
… | …0101101100010100110011 |
3 | 1201110200012111011100021001 |
4 | 2332332333111230110303 |
5 | 3210012114304132430 |
6 | 43525123015542431 |
7 | 2523126155514223 |
oct | 276767725542463 |
9 | 51420174140231 |
10 | 13124335224115 |
11 | 41aaaaa0a0965 |
12 | 157b70138aa17 |
13 | 742809075a4b |
14 | 335315024083 |
15 | 17b5da02d3ca |
hex | befbf56c533 |
13124335224115 has 8 divisors (see below), whose sum is σ = 15751289130480. Its totient is φ = 10498076938272.
The previous prime is 13124335224041. The next prime is 13124335224137. The reversal of 13124335224115 is 51142253342131.
It is a sphenic number, since it is the product of 3 distinct primes.
It is a cyclic number.
It is not a de Polignac number, because 13124335224115 - 213 = 13124335215923 is a prime.
It is an unprimeable number.
It is a pernicious number, because its binary representation contains a prime number (29) of ones.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 173863620 + ... + 173939089.
It is an arithmetic number, because the mean of its divisors is an integer number (1968911141310).
Almost surely, 213124335224115 is an apocalyptic number.
13124335224115 is a deficient number, since it is larger than the sum of its proper divisors (2626953906365).
13124335224115 is an equidigital number, since it uses as much as digits as its factorization.
13124335224115 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 347810261.
The product of its digits is 86400, while the sum is 37.
Adding to 13124335224115 its reverse (51142253342131), we get a palindrome (64266588566246).
The spelling of 13124335224115 in words is "thirteen trillion, one hundred twenty-four billion, three hundred thirty-five million, two hundred twenty-four thousand, one hundred fifteen".
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