Base | Representation |
---|---|
bin | 100000100010100100011101… |
… | …101011001001011010011111 |
3 | 200202201110221012222121211220 |
4 | 200202210131223021122133 |
5 | 122224231113344320111 |
6 | 1224213130230544423 |
7 | 42100403052144612 |
oct | 4042443553113237 |
9 | 622643835877756 |
10 | 143113103120031 |
11 | 41666a20840040 |
12 | 14074340808113 |
13 | 61b167a5b86a5 |
14 | 274a9d5574179 |
15 | 1182a7848ae06 |
hex | 82291dac969f |
143113103120031 has 32 divisors (see below), whose sum is σ = 208379624633856. Its totient is φ = 86645589744000.
The previous prime is 143113103120017. The next prime is 143113103120161. The reversal of 143113103120031 is 130021301311341.
It is not a de Polignac number, because 143113103120031 - 25 = 143113103119999 is a prime.
It is a super-2 number, since 2×1431131031200312 (a number of 29 digits) contains 22 as substring.
It is a junction number, because it is equal to n+sod(n) for n = 143113103119983 and 143113103120010.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (143113103820031) 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 31 ways as a sum of consecutive naturals, for example, 2127377926 + ... + 2127445196.
It is an arithmetic number, because the mean of its divisors is an integer number (6511863269808).
Almost surely, 2143113103120031 is an apocalyptic number.
143113103120031 is a gapful number since it is divisible by the number (11) formed by its first and last digit.
143113103120031 is a deficient number, since it is larger than the sum of its proper divisors (65266521513825).
143113103120031 is a wasteful number, since it uses less digits than its factorization.
143113103120031 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 132943.
The product of its (nonzero) digits is 648, while the sum is 24.
Adding to 143113103120031 its reverse (130021301311341), we get a palindrome (273134404431372).
The spelling of 143113103120031 in words is "one hundred forty-three trillion, one hundred thirteen billion, one hundred three million, one hundred twenty thousand, thirty-one".
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