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1304399989433 is a prime number
BaseRepresentation
bin10010111110110100010…
…011000100101010111001
311121200212212201002000202
4102332310103010222321
5132332402344130213
62435122202411545
7163145150654042
oct22766423045271
94550785632022
101304399989433
114632138743a2
12190974b00bb5
1396009855698
14471c1a9d8c9
1523de535ab58
hex12fb44c4ab9

1304399989433 has 2 divisors, whose sum is σ = 1304399989434. Its totient is φ = 1304399989432.

The previous prime is 1304399989369. The next prime is 1304399989439. The reversal of 1304399989433 is 3349899934031.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 1256076079504 + 48323909929 = 1120748^2 + 219827^2 .

It is a cyclic number.

It is not a de Polignac number, because 1304399989433 - 26 = 1304399989369 is a prime.

It is a super-2 number, since 2×13043999894332 (a number of 25 digits) contains 22 as substring.

It is a self number, because there is not a number n which added to its sum of digits gives 1304399989433.

It is not a weakly prime, because it can be changed into another prime (1304399989439) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 652199994716 + 652199994717.

It is an arithmetic number, because the mean of its divisors is an integer number (652199994717).

Almost surely, 21304399989433 is an apocalyptic number.

It is an amenable number.

1304399989433 is a deficient number, since it is larger than the sum of its proper divisors (1).

1304399989433 is an equidigital number, since it uses as much as digits as its factorization.

1304399989433 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 68024448, while the sum is 65.

The spelling of 1304399989433 in words is "one trillion, three hundred four billion, three hundred ninety-nine million, nine hundred eighty-nine thousand, four hundred thirty-three".