Base | Representation |
---|---|
bin | 111011111011101011… |
… | …011001100010000011 |
3 | 20011002210122022121021 |
4 | 323323223121202003 |
5 | 2023243103001011 |
6 | 45321330410311 |
7 | 4435462655506 |
oct | 737353314203 |
9 | 204083568537 |
10 | 64352000131 |
11 | 25323032191 |
12 | 1057b430997 |
13 | 60b72852b3 |
14 | 318687513d |
15 | 1a19859471 |
hex | efbad9883 |
64352000131 has 2 divisors, whose sum is σ = 64352000132. Its totient is φ = 64352000130.
The previous prime is 64352000099. The next prime is 64352000143. The reversal of 64352000131 is 13100025346.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 64352000131 - 25 = 64352000099 is a prime.
It is a super-2 number, since 2×643520001312 (a number of 22 digits) contains 22 as substring.
It is a junction number, because it is equal to n+sod(n) for n = 64352000096 and 64352000105.
It is not a weakly prime, because it can be changed into another prime (64352010131) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 32176000065 + 32176000066.
It is an arithmetic number, because the mean of its divisors is an integer number (32176000066).
Almost surely, 264352000131 is an apocalyptic number.
64352000131 is a deficient number, since it is larger than the sum of its proper divisors (1).
64352000131 is an equidigital number, since it uses as much as digits as its factorization.
64352000131 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 2160, while the sum is 25.
Adding to 64352000131 its reverse (13100025346), we get a palindrome (77452025477).
The spelling of 64352000131 in words is "sixty-four billion, three hundred fifty-two million, one hundred thirty-one", and thus it is an aban number.
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