Base | Representation |
---|---|
bin | 1101000010111000010000… |
… | …1010010111010111011011 |
3 | 1212210012002012202221110122 |
4 | 3100232010022113113123 |
5 | 3334444143424023003 |
6 | 50301045111503455 |
7 | 3010153466566205 |
oct | 320560412272733 |
9 | 55705065687418 |
10 | 14343113111003 |
11 | 462a975477336 |
12 | 173795b687b8b |
13 | 80071baa0a11 |
14 | 3782d33c9775 |
15 | 19d16d52c238 |
hex | d0b842975db |
14343113111003 has 4 divisors (see below), whose sum is σ = 14343194642208. Its totient is φ = 14343031579800.
The previous prime is 14343113110973. The next prime is 14343113111071. The reversal of 14343113111003 is 30011131134341.
14343113111003 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
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 14343113111003 - 232 = 14338818143707 is a prime.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (14343113101003) 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, 40501148 + ... + 40853753.
It is an arithmetic number, because the mean of its divisors is an integer number (3585798660552).
Almost surely, 214343113111003 is an apocalyptic number.
14343113111003 is a deficient number, since it is larger than the sum of its proper divisors (81531205).
14343113111003 is an equidigital number, since it uses as much as digits as its factorization.
14343113111003 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 81531204.
The product of its (nonzero) digits is 1296, while the sum is 26.
Adding to 14343113111003 its reverse (30011131134341), we get a palindrome (44354244245344).
The spelling of 14343113111003 in words is "fourteen trillion, three hundred forty-three billion, one hundred thirteen million, one hundred eleven thousand, three".
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