Base | Representation |
---|---|
bin | 10001011101101100001… |
… | …000010101100000001111 |
3 | 11020201200201200200210211 |
4 | 101131230020111200033 |
5 | 124130311130001033 |
6 | 2315153430143251 |
7 | 152463562455115 |
oct | 21355410254017 |
9 | 4221621620724 |
10 | 1200110000143 |
11 | 422a668988aa |
12 | 17470a4b1b27 |
13 | 89229320152 |
14 | 4212ad9dab5 |
15 | 2133e6b82cd |
hex | 1176c21580f |
1200110000143 has 2 divisors, whose sum is σ = 1200110000144. Its totient is φ = 1200110000142.
The previous prime is 1200110000107. The next prime is 1200110000153. The reversal of 1200110000143 is 3410000110021.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-1200110000143 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (1200110000153) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 600055000071 + 600055000072.
It is an arithmetic number, because the mean of its divisors is an integer number (600055000072).
Almost surely, 21200110000143 is an apocalyptic number.
1200110000143 is a deficient number, since it is larger than the sum of its proper divisors (1).
1200110000143 is an equidigital number, since it uses as much as digits as its factorization.
1200110000143 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 24, while the sum is 13.
Adding to 1200110000143 its reverse (3410000110021), we get a palindrome (4610110110164).
The spelling of 1200110000143 in words is "one trillion, two hundred billion, one hundred ten million, one hundred forty-three", and thus it is an aban number.
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