Base | Representation |
---|---|
bin | 1010000010001010001001… |
… | …1000010111100101110111 |
3 | 1110001200001000020001222021 |
4 | 2200202202120113211313 |
5 | 2421222421240010312 |
6 | 35244042312014011 |
7 | 2216023131264433 |
oct | 240424230274567 |
9 | 43050030201867 |
10 | 11032200313207 |
11 | 3573803516907 |
12 | 12a2145691307 |
13 | 6204415ca218 |
14 | 2a1d65cdd9c3 |
15 | 141e8d562c07 |
hex | a08a2617977 |
11032200313207 has 2 divisors, whose sum is σ = 11032200313208. Its totient is φ = 11032200313206.
The previous prime is 11032200313133. The next prime is 11032200313219. The reversal of 11032200313207 is 70231300223011.
It is a happy number.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-11032200313207 is a prime.
It is a super-4 number, since 4×110322003132074 (a number of 53 digits) contains 4444 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (11032200313237) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5516100156603 + 5516100156604.
It is an arithmetic number, because the mean of its divisors is an integer number (5516100156604).
Almost surely, 211032200313207 is an apocalyptic number.
11032200313207 is a deficient number, since it is larger than the sum of its proper divisors (1).
11032200313207 is an equidigital number, since it uses as much as digits as its factorization.
11032200313207 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 1512, while the sum is 25.
Adding to 11032200313207 its reverse (70231300223011), we get a palindrome (81263500536218).
The spelling of 11032200313207 in words is "eleven trillion, thirty-two billion, two hundred million, three hundred thirteen thousand, two hundred seven".
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