Base | Representation |
---|---|
bin | 101111010111101111000… |
… | …001111101110011101001 |
3 | 102112012111111020002110201 |
4 | 233113233001331303221 |
5 | 411313323412404001 |
6 | 10531244151312201 |
7 | 454121244545365 |
oct | 57275701756351 |
9 | 12465444202421 |
10 | 3255300513001 |
11 | 1045624301889 |
12 | 446a9567b661 |
13 | 1a7c86a97b04 |
14 | b37b36742a5 |
15 | 59a27a75501 |
hex | 2f5ef07dce9 |
3255300513001 has 2 divisors, whose sum is σ = 3255300513002. Its totient is φ = 3255300513000.
The previous prime is 3255300512969. The next prime is 3255300513013. The reversal of 3255300513001 is 1003150035523.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 3165111413776 + 90189099225 = 1779076^2 + 300315^2 .
It is a cyclic number.
It is not a de Polignac number, because 3255300513001 - 25 = 3255300512969 is a prime.
It is not a weakly prime, because it can be changed into another prime (3255300013001) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1627650256500 + 1627650256501.
It is an arithmetic number, because the mean of its divisors is an integer number (1627650256501).
Almost surely, 23255300513001 is an apocalyptic number.
It is an amenable number.
3255300513001 is a deficient number, since it is larger than the sum of its proper divisors (1).
3255300513001 is an equidigital number, since it uses as much as digits as its factorization.
3255300513001 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 6750, while the sum is 28.
Adding to 3255300513001 its reverse (1003150035523), we get a palindrome (4258450548524).
The spelling of 3255300513001 in words is "three trillion, two hundred fifty-five billion, three hundred million, five hundred thirteen thousand, one".
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