Base | Representation |
---|---|
bin | 1110011110101111000… |
… | …01000010001011000011 |
3 | 1202120020012122100121011 |
4 | 13033113201002023003 |
5 | 31122423433001011 |
6 | 1020322040341351 |
7 | 50642300502613 |
oct | 7172741021303 |
9 | 1676205570534 |
10 | 497537000131 |
11 | 182005736216 |
12 | 80514174857 |
13 | 37bc0aa48c2 |
14 | 1a11c084a43 |
15 | ce1e7b5621 |
hex | 73d78422c3 |
497537000131 has 2 divisors, whose sum is σ = 497537000132. Its totient is φ = 497537000130.
The previous prime is 497537000111. The next prime is 497537000143. The reversal of 497537000131 is 131000735794.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 497537000131 - 27 = 497537000003 is a prime.
It is a super-2 number, since 2×4975370001312 (a number of 24 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (497537000101) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (19) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 248768500065 + 248768500066.
It is an arithmetic number, because the mean of its divisors is an integer number (248768500066).
Almost surely, 2497537000131 is an apocalyptic number.
497537000131 is a deficient number, since it is larger than the sum of its proper divisors (1).
497537000131 is an equidigital number, since it uses as much as digits as its factorization.
497537000131 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 79380, while the sum is 40.
The spelling of 497537000131 in words is "four hundred ninety-seven billion, five hundred thirty-seven million, one hundred thirty-one", and thus it is an aban number.
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