Base | Representation |
---|---|
bin | 10010111010111010101… |
… | …000100010101001011001 |
3 | 11121022001221222021111111 |
4 | 102322322220202221121 |
5 | 132300313044010431 |
6 | 2433150412500321 |
7 | 162636306545101 |
oct | 22727250425131 |
9 | 4538057867444 |
10 | 1300211313241 |
11 | 461464431a35 |
12 | 18bba617a6a1 |
13 | 957bcb12c69 |
14 | 46d05674201 |
15 | 23c4c7667b1 |
hex | 12ebaa22a59 |
1300211313241 has 2 divisors, whose sum is σ = 1300211313242. Its totient is φ = 1300211313240.
The previous prime is 1300211313223. The next prime is 1300211313371. The reversal of 1300211313241 is 1423131120031.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 812314651225 + 487896662016 = 901285^2 + 698496^2 .
It is a cyclic number.
It is not a de Polignac number, because 1300211313241 - 25 = 1300211313209 is a prime.
It is not a weakly prime, because it can be changed into another prime (1300211313041) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 650105656620 + 650105656621.
It is an arithmetic number, because the mean of its divisors is an integer number (650105656621).
Almost surely, 21300211313241 is an apocalyptic number.
It is an amenable number.
1300211313241 is a deficient number, since it is larger than the sum of its proper divisors (1).
1300211313241 is an equidigital number, since it uses as much as digits as its factorization.
1300211313241 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 432, while the sum is 22.
Adding to 1300211313241 its reverse (1423131120031), we get a palindrome (2723342433272).
The spelling of 1300211313241 in words is "one trillion, three hundred billion, two hundred eleven million, three hundred thirteen thousand, two hundred forty-one".
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