Base | Representation |
---|---|
bin | 10001101001101110110… |
… | …101100100011011000011 |
3 | 11021222001212100200220221 |
4 | 101221232311210123003 |
5 | 124333301310022201 |
6 | 2325132451443511 |
7 | 153432161354236 |
oct | 21515665443303 |
9 | 4258055320827 |
10 | 1213040314051 |
11 | 4284a16a7045 |
12 | 1771188b6597 |
13 | 8a50a133177 |
14 | 429d638941d |
15 | 218499524a1 |
hex | 11a6ed646c3 |
1213040314051 has 2 divisors, whose sum is σ = 1213040314052. Its totient is φ = 1213040314050.
The previous prime is 1213040313977. The next prime is 1213040314067. The reversal of 1213040314051 is 1504130403121.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-1213040314051 is a prime.
It is not a weakly prime, because it can be changed into another prime (1213040314951) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 606520157025 + 606520157026.
It is an arithmetic number, because the mean of its divisors is an integer number (606520157026).
Almost surely, 21213040314051 is an apocalyptic number.
1213040314051 is a deficient number, since it is larger than the sum of its proper divisors (1).
1213040314051 is an equidigital number, since it uses as much as digits as its factorization.
1213040314051 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 1440, while the sum is 25.
Adding to 1213040314051 its reverse (1504130403121), we get a palindrome (2717170717172).
The spelling of 1213040314051 in words is "one trillion, two hundred thirteen billion, forty million, three hundred fourteen thousand, fifty-one".
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