Base | Representation |
---|---|
bin | 1000010001101100011110… |
… | …1100010000010100110001 |
3 | 1012012221220222200110021121 |
4 | 2010123013230100110301 |
5 | 2143043441244000423 |
6 | 31204305112022241 |
7 | 1626313511143231 |
oct | 204330754202461 |
9 | 35187828613247 |
10 | 9100091000113 |
11 | 2999366148a33 |
12 | 102b7aa091981 |
13 | 51019a4b1649 |
14 | 236637c73cc1 |
15 | 10baaa8de45d |
hex | 846c7b10531 |
9100091000113 has 2 divisors, whose sum is σ = 9100091000114. Its totient is φ = 9100091000112.
The previous prime is 9100091000101. The next prime is 9100091000129. The reversal of 9100091000113 is 3110001900019.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 7948948694544 + 1151142305569 = 2819388^2 + 1072913^2 .
It is a cyclic number.
It is not a de Polignac number, because 9100091000113 - 217 = 9100090869041 is a prime.
It is not a weakly prime, because it can be changed into another prime (9100091000153) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 4550045500056 + 4550045500057.
It is an arithmetic number, because the mean of its divisors is an integer number (4550045500057).
Almost surely, 29100091000113 is an apocalyptic number.
It is an amenable number.
9100091000113 is a deficient number, since it is larger than the sum of its proper divisors (1).
9100091000113 is an equidigital number, since it uses as much as digits as its factorization.
9100091000113 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 243, while the sum is 25.
The spelling of 9100091000113 in words is "nine trillion, one hundred billion, ninety-one million, one hundred thirteen", and thus it is an aban number.
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