Base | Representation |
---|---|
bin | 11000101101011000111001… |
… | …01000100010100100111011 |
3 | 21010101111000001012221102112 |
4 | 30112230130220202210323 |
5 | 24110220324040314011 |
6 | 311321355354322535 |
7 | 14305440333343412 |
oct | 1426543450424473 |
9 | 233344001187375 |
10 | 54336111651131 |
11 | 163498a753aa32 |
12 | 6116851a2944b |
13 | 2441b47970bcb |
14 | d5bc4ba88479 |
15 | 64361874258b |
hex | 316b1ca2293b |
54336111651131 has 2 divisors, whose sum is σ = 54336111651132. Its totient is φ = 54336111651130.
The previous prime is 54336111651119. The next prime is 54336111651139. The reversal of 54336111651131 is 13115611163345.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 54336111651131 - 230 = 54335037909307 is a prime.
It is a super-2 number, since 2×543361116511312 (a number of 28 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (54336111651139) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 27168055825565 + 27168055825566.
It is an arithmetic number, because the mean of its divisors is an integer number (27168055825566).
Almost surely, 254336111651131 is an apocalyptic number.
54336111651131 is a deficient number, since it is larger than the sum of its proper divisors (1).
54336111651131 is an equidigital number, since it uses as much as digits as its factorization.
54336111651131 is an evil number, because the sum of its binary digits is even.
The product of its digits is 97200, while the sum is 41.
The spelling of 54336111651131 in words is "fifty-four trillion, three hundred thirty-six billion, one hundred eleven million, six hundred fifty-one thousand, one hundred thirty-one".
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