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804939939413 is a prime number
BaseRepresentation
bin10111011011010100010…
…10001011111001010101
32211221200122111122220122
423231222022023321111
5101142004111030123
61413441224241325
7112104106410521
oct13555212137125
92757618448818
10804939939413
1129041167a355
12110004910245
135aba053984c
142ad60470d81
1515e11d32bc8
hexbb6a28be55

804939939413 has 2 divisors, whose sum is σ = 804939939414. Its totient is φ = 804939939412.

The previous prime is 804939939367. The next prime is 804939939421. The reversal of 804939939413 is 314939939408.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 625755356209 + 179184583204 = 791047^2 + 423302^2 .

It is a cyclic number.

It is not a de Polignac number, because 804939939413 - 228 = 804671503957 is a prime.

It is a super-2 number, since 2×8049399394132 (a number of 25 digits) contains 22 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (804939939313) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 402469969706 + 402469969707.

It is an arithmetic number, because the mean of its divisors is an integer number (402469969707).

Almost surely, 2804939939413 is an apocalyptic number.

It is an amenable number.

804939939413 is a deficient number, since it is larger than the sum of its proper divisors (1).

804939939413 is an equidigital number, since it uses as much as digits as its factorization.

804939939413 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 22674816, while the sum is 62.

The spelling of 804939939413 in words is "eight hundred four billion, nine hundred thirty-nine million, nine hundred thirty-nine thousand, four hundred thirteen".