Base | Representation |
---|---|
bin | 1001001100100010110100… |
… | …0111101010110111100001 |
3 | 1022210121112122022002002011 |
4 | 2103020231013222313201 |
5 | 2311130023130000342 |
6 | 33300551345552521 |
7 | 2062334505640531 |
oct | 223105507526741 |
9 | 38717478262064 |
10 | 10111110000097 |
11 | 3249108aa9687 |
12 | 1173726345741 |
13 | 5846205c7a02 |
14 | 26d5478d54c1 |
15 | 12802e6e9717 |
hex | 9322d1eade1 |
10111110000097 has 2 divisors, whose sum is σ = 10111110000098. Its totient is φ = 10111110000096.
The previous prime is 10111110000091. The next prime is 10111110000133. The reversal of 10111110000097 is 79000001111101.
10111110000097 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 8993017354896 + 1118092645201 = 2998836^2 + 1057399^2 .
It is a cyclic number.
It is not a de Polignac number, because 10111110000097 - 211 = 10111109998049 is a prime.
It is not a weakly prime, because it can be changed into another prime (10111110000091) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5055555000048 + 5055555000049.
It is an arithmetic number, because the mean of its divisors is an integer number (5055555000049).
Almost surely, 210111110000097 is an apocalyptic number.
It is an amenable number.
10111110000097 is a deficient number, since it is larger than the sum of its proper divisors (1).
10111110000097 is an equidigital number, since it uses as much as digits as its factorization.
10111110000097 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 63, while the sum is 22.
Adding to 10111110000097 its reverse (79000001111101), we get a palindrome (89111111111198).
The spelling of 10111110000097 in words is "ten trillion, one hundred eleven billion, one hundred ten million, ninety-seven", and thus it is an aban number.
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