Base | Representation |
---|---|
bin | 11101110100010111100110… |
… | …011101100110010101011011 |
3 | 122012100001200122020102021112 |
4 | 131310113212131212111123 |
5 | 114142113011324431033 |
6 | 1142525503404130535 |
7 | 36423463364564015 |
oct | 3564274635462533 |
9 | 565301618212245 |
10 | 131142102967643 |
11 | 38871065291a49 |
12 | 12860274b36a4b |
13 | 5823854ccc561 |
14 | 24554512531b5 |
15 | 1026491482c48 |
hex | 7745e676655b |
131142102967643 has 2 divisors, whose sum is σ = 131142102967644. Its totient is φ = 131142102967642.
The previous prime is 131142102967591. The next prime is 131142102967669. The reversal of 131142102967643 is 346769201241131.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 131142102967643 - 210 = 131142102966619 is a prime.
It is a Sophie Germain prime.
It is a junction number, because it is equal to n+sod(n) for n = 131142102967591 and 131142102967600.
It is not a weakly prime, because it can be changed into another prime (131142102967243) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65571051483821 + 65571051483822.
It is an arithmetic number, because the mean of its divisors is an integer number (65571051483822).
Almost surely, 2131142102967643 is an apocalyptic number.
131142102967643 is a deficient number, since it is larger than the sum of its proper divisors (1).
131142102967643 is an equidigital number, since it uses as much as digits as its factorization.
131142102967643 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 1306368, while the sum is 50.
The spelling of 131142102967643 in words is "one hundred thirty-one trillion, one hundred forty-two billion, one hundred two million, nine hundred sixty-seven thousand, six hundred forty-three".
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