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13531443030503 = 22376048924019
BaseRepresentation
bin1100010011101000100011…
…0110101110100111100111
31202220121000112022200121022
43010322020312232213213
53233144343403434003
644440131545041355
72564420526001646
oct304721066564747
952817015280538
1013531443030503
11434771a910412
121626598a8625b
13772018134a88
1434acd51a735d
15186eb5938738
hexc4e88dae9e7

13531443030503 has 4 divisors (see below), whose sum is σ = 13537491956760. Its totient is φ = 13525394104248.

The previous prime is 13531443030499. The next prime is 13531443030569. The reversal of 13531443030503 is 30503034413531.

It is a happy number.

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 13531443030503 - 22 = 13531443030499 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 (13531443030583) 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, 3024459773 + ... + 3024464246.

It is an arithmetic number, because the mean of its divisors is an integer number (3384372989190).

Almost surely, 213531443030503 is an apocalyptic number.

13531443030503 is a deficient number, since it is larger than the sum of its proper divisors (6048926257).

13531443030503 is an equidigital number, since it uses as much as digits as its factorization.

13531443030503 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 6048926256.

The product of its (nonzero) digits is 97200, while the sum is 35.

The spelling of 13531443030503 in words is "thirteen trillion, five hundred thirty-one billion, four hundred forty-three million, thirty thousand, five hundred three".

Divisors: 1 2237 6048924019 13531443030503