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13104332552551 = 31422720404921
BaseRepresentation
bin1011111010110001011100…
…0101100011100101100111
31201101202120110121100022111
42332230113011203211213
53204200133113140201
643512014113141451
72521520401562422
oct276542705434547
951352513540274
1013104332552551
1141a2575107739
12157785a55a887
1374096bc74106
143343786449b9
1517ad18e35b51
hexbeb17163967

13104332552551 has 4 divisors (see below), whose sum is σ = 13527052957504. Its totient is φ = 12681612147600.

The previous prime is 13104332552537. The next prime is 13104332552573. The reversal of 13104332552551 is 15525523340131.

It is a semiprime because it is the product of two primes.

It is a cyclic number.

It is not a de Polignac number, because 13104332552551 - 233 = 13095742617959 is a prime.

It is a Duffinian number.

It is a congruent number.

It is not an unprimeable number, because it can be changed into a prime (13104332552581) by changing a digit.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 211360202430 + ... + 211360202491.

It is an arithmetic number, because the mean of its divisors is an integer number (3381763239376).

Almost surely, 213104332552551 is an apocalyptic number.

13104332552551 is a deficient number, since it is larger than the sum of its proper divisors (422720404953).

13104332552551 is an equidigital number, since it uses as much as digits as its factorization.

13104332552551 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 422720404952.

The product of its (nonzero) digits is 270000, while the sum is 40.

Adding to 13104332552551 its reverse (15525523340131), we get a palindrome (28629855892682).

The spelling of 13104332552551 in words is "thirteen trillion, one hundred four billion, three hundred thirty-two million, five hundred fifty-two thousand, five hundred fifty-one".

Divisors: 1 31 422720404921 13104332552551