Base | Representation |
---|---|
bin | 1010000000011010001011… |
… | …1100011101001101000011 |
3 | 1102221210110101212002022022 |
4 | 2200012202330131031003 |
5 | 2420224343110033212 |
6 | 35222152102232055 |
7 | 2213610260331344 |
oct | 240064274351503 |
9 | 42853411762268 |
10 | 11002145002307 |
11 | 3561a90034998 |
12 | 129835812562b |
13 | 61a66193a5a4 |
14 | 2a07143b80cb |
15 | 1412ceb60772 |
hex | a01a2f1d343 |
11002145002307 has 2 divisors, whose sum is σ = 11002145002308. Its totient is φ = 11002145002306.
The previous prime is 11002145002291. The next prime is 11002145002319. The reversal of 11002145002307 is 70320054120011.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 11002145002307 - 24 = 11002145002291 is a prime.
It is not a weakly prime, because it can be changed into another prime (11002145002357) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (19) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5501072501153 + 5501072501154.
It is an arithmetic number, because the mean of its divisors is an integer number (5501072501154).
Almost surely, 211002145002307 is an apocalyptic number.
11002145002307 is a deficient number, since it is larger than the sum of its proper divisors (1).
11002145002307 is an equidigital number, since it uses as much as digits as its factorization.
11002145002307 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 1680, while the sum is 26.
Adding to 11002145002307 its reverse (70320054120011), we get a palindrome (81322199122318).
The spelling of 11002145002307 in words is "eleven trillion, two billion, one hundred forty-five million, two thousand, three hundred seven".
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