Base | Representation |
---|---|
bin | 1001001011001010011… |
… | …11010101010110100101 |
3 | 1010010122222222122121202 |
4 | 10211211033111112211 |
5 | 20131043132143140 |
6 | 400452025052245 |
7 | 31526502255245 |
oct | 4454517252645 |
9 | 1103588878552 |
10 | 315231131045 |
11 | 111763834506 |
12 | 51116327085 |
13 | 2395951199c |
14 | 11385d2db25 |
15 | 82ee92b915 |
hex | 49653d55a5 |
315231131045 has 4 divisors (see below), whose sum is σ = 378277357260. Its totient is φ = 252184904832.
The previous prime is 315231131003. The next prime is 315231131071. The reversal of 315231131045 is 540131132513.
It is a semiprime because it is the product of two primes.
It can be written as a sum of positive squares in 2 ways, for example, as 78418481089 + 236812649956 = 280033^2 + 486634^2 .
It is a cyclic number.
It is not a de Polignac number, because 315231131045 - 212 = 315231126949 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 315231130999 and 315231131017.
It is a congruent number.
It is an unprimeable number.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 31523113100 + ... + 31523113109.
It is an arithmetic number, because the mean of its divisors is an integer number (94569339315).
Almost surely, 2315231131045 is an apocalyptic number.
It is an amenable number.
315231131045 is a deficient number, since it is larger than the sum of its proper divisors (63046226215).
315231131045 is an equidigital number, since it uses as much as digits as its factorization.
315231131045 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 63046226214.
The product of its (nonzero) digits is 5400, while the sum is 29.
Adding to 315231131045 its reverse (540131132513), we get a palindrome (855362263558).
The spelling of 315231131045 in words is "three hundred fifteen billion, two hundred thirty-one million, one hundred thirty-one thousand, forty-five".
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