Base | Representation |
---|---|
bin | 11111000000111011011… |
… | …000101101001010011011 |
3 | 21112202020222011120121021 |
4 | 133000323120231022123 |
5 | 234404400312023003 |
6 | 4311034502023311 |
7 | 306660414014350 |
oct | 37007330551233 |
9 | 7482228146537 |
10 | 2131300111003 |
11 | 751974393618 |
12 | 2a5088117b37 |
13 | 125c998b254b |
14 | 752268a8627 |
15 | 3a6900542bd |
hex | 1f03b62d29b |
2131300111003 has 4 divisors (see below), whose sum is σ = 2435771555440. Its totient is φ = 1826828666568.
The previous prime is 2131300110983. The next prime is 2131300111013. The reversal of 2131300111003 is 3001110031312.
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-2131300111003 is a prime.
It is a super-2 number, since 2×21313001110032 (a number of 25 digits) contains 22 as substring.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (2131300111013) 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, 152235722208 + ... + 152235722221.
It is an arithmetic number, because the mean of its divisors is an integer number (608942888860).
Almost surely, 22131300111003 is an apocalyptic number.
2131300111003 is a deficient number, since it is larger than the sum of its proper divisors (304471444437).
2131300111003 is an equidigital number, since it uses as much as digits as its factorization.
2131300111003 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 304471444436.
The product of its (nonzero) digits is 54, while the sum is 16.
Adding to 2131300111003 its reverse (3001110031312), we get a palindrome (5132410142315).
The spelling of 2131300111003 in words is "two trillion, one hundred thirty-one billion, three hundred million, one hundred eleven thousand, three".
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