Base | Representation |
---|---|
bin | 11110000000100111001… |
… | …10111101100111110001 |
3 | 10122120111121022201212012 |
4 | 33000103212331213301 |
5 | 113343214002211032 |
6 | 2105405052212305 |
7 | 134332100003234 |
oct | 17002346754761 |
9 | 3576447281765 |
10 | 1031121132017 |
11 | 368328aa8009 |
12 | 147a0846a095 |
13 | 76307a00062 |
14 | 37c9973141b |
15 | 1bc4d9ec6b2 |
hex | f0139bd9f1 |
1031121132017 has 2 divisors, whose sum is σ = 1031121132018. Its totient is φ = 1031121132016.
The previous prime is 1031121132007. The next prime is 1031121132049. The reversal of 1031121132017 is 7102311211301.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 895173330496 + 135947801521 = 946136^2 + 368711^2 .
It is a cyclic number.
It is not a de Polignac number, because 1031121132017 - 234 = 1013941262833 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 1031121131983 and 1031121132001.
It is not a weakly prime, because it can be changed into another prime (1031121132007) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 515560566008 + 515560566009.
It is an arithmetic number, because the mean of its divisors is an integer number (515560566009).
Almost surely, 21031121132017 is an apocalyptic number.
It is an amenable number.
1031121132017 is a deficient number, since it is larger than the sum of its proper divisors (1).
1031121132017 is an equidigital number, since it uses as much as digits as its factorization.
1031121132017 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 252, while the sum is 23.
Adding to 1031121132017 its reverse (7102311211301), we get a palindrome (8133432343318).
The spelling of 1031121132017 in words is "one trillion, thirty-one billion, one hundred twenty-one million, one hundred thirty-two thousand, seventeen".
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