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33320134845793 is a prime number
BaseRepresentation
bin1111001001101111100101…
…10011010010010101100001
311100222101002111011012120201
413210313302303102111201
513331404114010031133
6154511020355305201
710006205001055165
oct744676263222541
9140871074135521
1033320134845793
11a686a98514754
1238a17b98ab201
1315790c9399993
148329bb56cca5
153cbaee6d1d7d
hex1e4df2cd2561

33320134845793 has 2 divisors, whose sum is σ = 33320134845794. Its totient is φ = 33320134845792.

The previous prime is 33320134845773. The next prime is 33320134845827. The reversal of 33320134845793 is 39754843102333.

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., 31502985013504 + 1817149832289 = 5612752^2 + 1348017^2 .

It is a cyclic number.

It is not a de Polignac number, because 33320134845793 - 25 = 33320134845761 is a prime.

It is a super-2 number, since 2×333201348457932 (a number of 28 digits) contains 22 as substring.

It is not a weakly prime, because it can be changed into another prime (33320134845733) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 16660067422896 + 16660067422897.

It is an arithmetic number, because the mean of its divisors is an integer number (16660067422897).

Almost surely, 233320134845793 is an apocalyptic number.

It is an amenable number.

33320134845793 is a deficient number, since it is larger than the sum of its proper divisors (1).

33320134845793 is an equidigital number, since it uses as much as digits as its factorization.

33320134845793 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 19595520, while the sum is 55.

The spelling of 33320134845793 in words is "thirty-three trillion, three hundred twenty billion, one hundred thirty-four million, eight hundred forty-five thousand, seven hundred ninety-three".