Base | Representation |
---|---|
bin | 11101110100100001111000… |
… | …100111101010100000010011 |
3 | 122012101010012011210111200012 |
4 | 131310201320213222200103 |
5 | 114142303120110424131 |
6 | 1142534531213345135 |
7 | 36424334124460601 |
oct | 3564417047524023 |
9 | 565333164714605 |
10 | 131153145014291 |
11 | 38875811229034 |
12 | 12862436a981ab |
13 | 58248c485917a |
14 | 2455bbb94a871 |
15 | 10268daa7a52b |
hex | 7748789ea813 |
131153145014291 has 2 divisors, whose sum is σ = 131153145014292. Its totient is φ = 131153145014290.
The previous prime is 131153145014267. The next prime is 131153145014299. The reversal of 131153145014291 is 192410541351131.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 131153145014291 - 226 = 131153077905427 is a prime.
It is a super-2 number, since 2×1311531450142912 (a number of 29 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (131153145014299) 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 as a sum of consecutive naturals, namely, 65576572507145 + 65576572507146.
It is an arithmetic number, because the mean of its divisors is an integer number (65576572507146).
Almost surely, 2131153145014291 is an apocalyptic number.
131153145014291 is a deficient number, since it is larger than the sum of its proper divisors (1).
131153145014291 is an equidigital number, since it uses as much as digits as its factorization.
131153145014291 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 64800, while the sum is 41.
The spelling of 131153145014291 in words is "one hundred thirty-one trillion, one hundred fifty-three billion, one hundred forty-five million, fourteen thousand, two hundred ninety-one".
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