Base | Representation |
---|---|
bin | 11101110100010110101010… |
… | …101011100101010010010011 |
3 | 122012022222002201222101212201 |
4 | 131310112222223211102103 |
5 | 114142103433100003103 |
6 | 1142525220055104031 |
7 | 36423426465524203 |
oct | 3564265253452223 |
9 | 565288081871781 |
10 | 131141100000403 |
11 | 388706a01218a9 |
12 | 12860035072017 |
13 | 582372529390c |
14 | 2455397d72403 |
15 | 10264333b751d |
hex | 7745aaae5493 |
131141100000403 has 2 divisors, whose sum is σ = 131141100000404. Its totient is φ = 131141100000402.
The previous prime is 131141100000353. The next prime is 131141100000431. The reversal of 131141100000403 is 304000001141131.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-131141100000403 is a prime.
It is not a weakly prime, because it can be changed into another prime (131141100000103) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65570550000201 + 65570550000202.
It is an arithmetic number, because the mean of its divisors is an integer number (65570550000202).
Almost surely, 2131141100000403 is an apocalyptic number.
131141100000403 is a deficient number, since it is larger than the sum of its proper divisors (1).
131141100000403 is an equidigital number, since it uses as much as digits as its factorization.
131141100000403 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 144, while the sum is 19.
Adding to 131141100000403 its reverse (304000001141131), we get a palindrome (435141101141534).
The spelling of 131141100000403 in words is "one hundred thirty-one trillion, one hundred forty-one billion, one hundred million, four hundred three", and thus it is an aban number.
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