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133007025362747 is a prime number
BaseRepresentation
bin11110001111100000011100…
…011111111101011100111011
3122102221100101021201222011012
4132033200130133331130323
5114413141341443101442
61150514330043224135
740005304036440644
oct3617403437753473
9572840337658135
10133007025362747
113941aa65370a18
1212b0179b81804b
13592a67c72b363
1424bb8081346cb
151059c40966582
hex78f81c7fd73b

133007025362747 has 2 divisors, whose sum is σ = 133007025362748. Its totient is φ = 133007025362746.

The previous prime is 133007025362737. The next prime is 133007025362819. The reversal of 133007025362747 is 747263520700331.

It is a weak prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-133007025362747 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 133007025362695 and 133007025362704.

It is not a weakly prime, because it can be changed into another prime (133007025362737) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 66503512681373 + 66503512681374.

It is an arithmetic number, because the mean of its divisors is an integer number (66503512681374).

Almost surely, 2133007025362747 is an apocalyptic number.

133007025362747 is a deficient number, since it is larger than the sum of its proper divisors (1).

133007025362747 is an equidigital number, since it uses as much as digits as its factorization.

133007025362747 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 4445280, while the sum is 50.

The spelling of 133007025362747 in words is "one hundred thirty-three trillion, seven billion, twenty-five million, three hundred sixty-two thousand, seven hundred forty-seven".