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133201324401049 is a prime number
BaseRepresentation
bin11110010010010101011001…
…100111111110000110011001
3122110121220220022122210102001
4132102111121213332012121
5114424332303021313144
61151143502122041001
740025322664142242
oct3622253147760631
9573556808583361
10133201324401049
1139495401089223
1212b33386049161
135942aa59038c8
1424c6d9b009cc9
15105ed13671bd4
hex7925599fe199

133201324401049 has 2 divisors, whose sum is σ = 133201324401050. Its totient is φ = 133201324401048.

The previous prime is 133201324401007. The next prime is 133201324401127. The reversal of 133201324401049 is 940104423102331.

It is a happy number.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 105994334576025 + 27206989825024 = 10295355^2 + 5216032^2 .

It is an emirp because it is prime and its reverse (940104423102331) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 133201324401049 - 221 = 133201322303897 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 133201324400999 and 133201324401017.

It is not a weakly prime, because it can be changed into another prime (133201324401749) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 66600662200524 + 66600662200525.

It is an arithmetic number, because the mean of its divisors is an integer number (66600662200525).

Almost surely, 2133201324401049 is an apocalyptic number.

It is an amenable number.

133201324401049 is a deficient number, since it is larger than the sum of its proper divisors (1).

133201324401049 is an equidigital number, since it uses as much as digits as its factorization.

133201324401049 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 62208, while the sum is 37.

The spelling of 133201324401049 in words is "one hundred thirty-three trillion, two hundred one billion, three hundred twenty-four million, four hundred one thousand, forty-nine".