Base | Representation |
---|---|
bin | 10011000101010110111… |
… | …011011001011101101011 |
3 | 11122101000100120011100111 |
4 | 103011112323121131223 |
5 | 132441243344433101 |
6 | 2442243135315151 |
7 | 163514203200406 |
oct | 23052673313553 |
9 | 4571010504314 |
10 | 1311423436651 |
11 | 466198383229 |
12 | 1921b507bab7 |
13 | 9688897ab03 |
14 | 4768a7a0d3d |
15 | 241a6c57251 |
hex | 13156ed976b |
1311423436651 has 4 divisors (see below), whose sum is σ = 1341921656152. Its totient is φ = 1280925217152.
The previous prime is 1311423436633. The next prime is 1311423436657. The reversal of 1311423436651 is 1566343241131.
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 1311423436651 - 223 = 1311415048043 is a prime.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (1311423436657) 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, 15249109686 + ... + 15249109771.
It is an arithmetic number, because the mean of its divisors is an integer number (335480414038).
Almost surely, 21311423436651 is an apocalyptic number.
1311423436651 is a deficient number, since it is larger than the sum of its proper divisors (30498219501).
1311423436651 is an equidigital number, since it uses as much as digits as its factorization.
1311423436651 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 30498219500.
The product of its digits is 155520, while the sum is 40.
Adding to 1311423436651 its reverse (1566343241131), we get a palindrome (2877766677782).
The spelling of 1311423436651 in words is "one trillion, three hundred eleven billion, four hundred twenty-three million, four hundred thirty-six thousand, six hundred fifty-one".
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