Base | Representation |
---|---|
bin | 11101011100111111110… |
… | …10000000111100110011 |
3 | 10120202011000201200012122 |
4 | 32232133322000330303 |
5 | 113040034011201042 |
6 | 2052523442144455 |
7 | 133054206630521 |
oct | 16563772007463 |
9 | 3522130650178 |
10 | 1012000100147 |
11 | 360206745833 |
12 | 14417095212b |
13 | 7457b466b0a |
14 | 36da408d311 |
15 | 1b4d00067d2 |
hex | eb9fe80f33 |
1012000100147 has 2 divisors, whose sum is σ = 1012000100148. Its totient is φ = 1012000100146.
The previous prime is 1012000100099. The next prime is 1012000100149. The reversal of 1012000100147 is 7410010002101.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 1012000100147 - 214 = 1012000083763 is a prime.
Together with 1012000100149, it forms a pair of twin primes.
It is a Chen prime.
It is not a weakly prime, because it can be changed into another prime (1012000100149) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 506000050073 + 506000050074.
It is an arithmetic number, because the mean of its divisors is an integer number (506000050074).
Almost surely, 21012000100147 is an apocalyptic number.
1012000100147 is a deficient number, since it is larger than the sum of its proper divisors (1).
1012000100147 is an equidigital number, since it uses as much as digits as its factorization.
1012000100147 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 56, while the sum is 17.
Adding to 1012000100147 its reverse (7410010002101), we get a palindrome (8422010102248).
The spelling of 1012000100147 in words is "one trillion, twelve billion, one hundred thousand, one hundred forty-seven".
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