Base | Representation |
---|---|
bin | 1100010101000001001110… |
… | …0000100111011011101011 |
3 | 1202222212110021020110121211 |
4 | 3011100103200213123223 |
5 | 3234042114442302103 |
6 | 44455105420413551 |
7 | 2566222404534022 |
oct | 305202340473353 |
9 | 52885407213554 |
10 | 13555244103403 |
11 | 4356823a037a2 |
12 | 162b11b9a72b7 |
13 | 77433c161272 |
14 | 34c11221d8b9 |
15 | 18790a23336d |
hex | c54138276eb |
13555244103403 has 4 divisors (see below), whose sum is σ = 13563807114528. Its totient is φ = 13546681092280.
The previous prime is 13555244103341. The next prime is 13555244103407. The reversal of 13555244103403 is 30430144255531.
It is a semiprime because it is the product of two primes.
It is a cyclic number.
It is not a de Polignac number, because 13555244103403 - 217 = 13555243972331 is a prime.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (13555244103407) by changing a digit.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 4281503188 + ... + 4281506353.
It is an arithmetic number, because the mean of its divisors is an integer number (3390951778632).
Almost surely, 213555244103403 is an apocalyptic number.
13555244103403 is a deficient number, since it is larger than the sum of its proper divisors (8563011125).
13555244103403 is an equidigital number, since it uses as much as digits as its factorization.
13555244103403 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 8563011124.
The product of its (nonzero) digits is 432000, while the sum is 40.
Adding to 13555244103403 its reverse (30430144255531), we get a palindrome (43985388358934).
The spelling of 13555244103403 in words is "thirteen trillion, five hundred fifty-five billion, two hundred forty-four million, one hundred three thousand, four hundred three".
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