Base | Representation |
---|---|
bin | 101101100000000001110111… |
… | …100000010101100110001011 |
3 | 222020112121221111011111020101 |
4 | 231200001313200111212023 |
5 | 202212123133013033003 |
6 | 1545334410533510231 |
7 | 60102460456655614 |
oct | 5540016740254613 |
9 | 866477844144211 |
10 | 200113121221003 |
11 | 58842552a99a51 |
12 | 1a53b31b4a2377 |
13 | 8787781303c03 |
14 | 375b753566c0b |
15 | 18206051c311d |
hex | b6007781598b |
200113121221003 has 2 divisors, whose sum is σ = 200113121221004. Its totient is φ = 200113121221002.
The previous prime is 200113121220997. The next prime is 200113121221007. The reversal of 200113121221003 is 300122121311002.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-200113121221003 is a prime.
It is a super-2 number, since 2×2001131212210032 (a number of 29 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (200113121221007) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 100056560610501 + 100056560610502.
It is an arithmetic number, because the mean of its divisors is an integer number (100056560610502).
Almost surely, 2200113121221003 is an apocalyptic number.
200113121221003 is a deficient number, since it is larger than the sum of its proper divisors (1).
200113121221003 is an equidigital number, since it uses as much as digits as its factorization.
200113121221003 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 144, while the sum is 19.
Adding to 200113121221003 its reverse (300122121311002), we get a palindrome (500235242532005).
The spelling of 200113121221003 in words is "two hundred trillion, one hundred thirteen billion, one hundred twenty-one million, two hundred twenty-one thousand, three".
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