Base | Representation |
---|---|
bin | 10110101111110001101101… |
… | …010111000001101011001011 |
3 | 111010012202021110012011010221 |
4 | 112233301231113001223023 |
5 | 101103024324012324042 |
6 | 552433501151050511 |
7 | 30033444511120633 |
oct | 2657615527015313 |
9 | 433182243164127 |
10 | 100040213011147 |
11 | 2996a8a8224692 |
12 | b278568b10a37 |
13 | 43a89a0389ab6 |
14 | 1a9bd877cb6c3 |
15 | b8741ed30d67 |
hex | 5afc6d5c1acb |
100040213011147 has 2 divisors, whose sum is σ = 100040213011148. Its totient is φ = 100040213011146.
The previous prime is 100040213011091. The next prime is 100040213011163. The reversal of 100040213011147 is 741110312040001.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-100040213011147 is a prime.
It is a super-3 number, since 3×1000402130111473 (a number of 43 digits) contains 333 as substring.
It is not a weakly prime, because it can be changed into another prime (100040213011747) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 50020106505573 + 50020106505574.
It is an arithmetic number, because the mean of its divisors is an integer number (50020106505574).
Almost surely, 2100040213011147 is an apocalyptic number.
100040213011147 is a deficient number, since it is larger than the sum of its proper divisors (1).
100040213011147 is an equidigital number, since it uses as much as digits as its factorization.
100040213011147 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 672, while the sum is 25.
Adding to 100040213011147 its reverse (741110312040001), we get a palindrome (841150525051148).
The spelling of 100040213011147 in words is "one hundred trillion, forty billion, two hundred thirteen million, eleven thousand, one hundred forty-seven".
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