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3155440311143 = 21117714942159
BaseRepresentation
bin101101111010101110111…
…001100001101101100111
3102011122202001122020012222
4231322232321201231213
5403144320204424033
610413331140502555
7443654516046551
oct55725671415547
912148661566188
103155440311143
11100723a88013a
1242b6666aaa5b
1319b731224883
14aca1d0044d1
1557130c52598
hex2deaee61b67

3155440311143 has 4 divisors (see below), whose sum is σ = 3155455464480. Its totient is φ = 3155425157808.

The previous prime is 3155440311109. The next prime is 3155440311157. The reversal of 3155440311143 is 3411130445513.

It is a happy number.

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

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-3155440311143 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 (3155440311173) 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, 7259903 + ... + 7682256.

It is an arithmetic number, because the mean of its divisors is an integer number (788863866120).

Almost surely, 23155440311143 is an apocalyptic number.

3155440311143 is a deficient number, since it is larger than the sum of its proper divisors (15153337).

3155440311143 is a wasteful number, since it uses less digits than its factorization.

3155440311143 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 15153336.

The product of its (nonzero) digits is 43200, while the sum is 35.

Adding to 3155440311143 its reverse (3411130445513), we get a palindrome (6566570756656).

The spelling of 3155440311143 in words is "three trillion, one hundred fifty-five billion, four hundred forty million, three hundred eleven thousand, one hundred forty-three".

Divisors: 1 211177 14942159 3155440311143