Base | Representation |
---|---|
bin | 101110011011100000… |
… | …101001001101010011 |
3 | 11202200101110122021221 |
4 | 232123200221031103 |
5 | 1304100014304011 |
6 | 34522533025511 |
7 | 3413264112253 |
oct | 563340511523 |
9 | 152611418257 |
10 | 49853666131 |
11 | 1a163105275 |
12 | 97b3a85297 |
13 | 4916655739 |
14 | 25ad11b963 |
15 | 146badb971 |
hex | b9b829353 |
49853666131 has 2 divisors, whose sum is σ = 49853666132. Its totient is φ = 49853666130.
The previous prime is 49853666113. The next prime is 49853666141. The reversal of 49853666131 is 13166635894.
49853666131 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
Together with previous prime (49853666113) it forms an Ormiston pair, because they use the same digits, order apart.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 49853666131 - 213 = 49853657939 is a prime.
It is a super-2 number, since 2×498536661312 (a number of 22 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (49853666141) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 24926833065 + 24926833066.
It is an arithmetic number, because the mean of its divisors is an integer number (24926833066).
Almost surely, 249853666131 is an apocalyptic number.
49853666131 is a deficient number, since it is larger than the sum of its proper divisors (1).
49853666131 is an equidigital number, since it uses as much as digits as its factorization.
49853666131 is an evil number, because the sum of its binary digits is even.
The product of its digits is 2799360, while the sum is 52.
The spelling of 49853666131 in words is "forty-nine billion, eight hundred fifty-three million, six hundred sixty-six thousand, one hundred thirty-one".
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