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12354787693 is a prime number
BaseRepresentation
bin10111000000110011…
…10001100101101101
31011220000201020120101
423200121301211231
5200300311201233
65401541551101
7615106651123
oct134031614555
934800636511
1012354787693
11526aa56946
122489715491
13121b805c81
14852bb3913
154c49a2c7d
hex2e067196d

12354787693 has 2 divisors, whose sum is σ = 12354787694. Its totient is φ = 12354787692.

The previous prime is 12354787661. The next prime is 12354787717. The reversal of 12354787693 is 39678745321.

12354787693 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 11611571049 + 743216644 = 107757^2 + 27262^2 .

It is a cyclic number.

It is not a de Polignac number, because 12354787693 - 25 = 12354787661 is a prime.

It is a super-3 number, since 3×123547876933 (a number of 31 digits) contains 333 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (12354781693) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (17) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6177393846 + 6177393847.

It is an arithmetic number, because the mean of its divisors is an integer number (6177393847).

Almost surely, 212354787693 is an apocalyptic number.

It is an amenable number.

12354787693 is a deficient number, since it is larger than the sum of its proper divisors (1).

12354787693 is an equidigital number, since it uses as much as digits as its factorization.

12354787693 is an odious number, because the sum of its binary digits is odd.

The product of its digits is 7620480, while the sum is 55.

The spelling of 12354787693 in words is "twelve billion, three hundred fifty-four million, seven hundred eighty-seven thousand, six hundred ninety-three".