Base | Representation |
---|---|
bin | 11111110101011100100010… |
… | …111000000011000111010111 |
3 | 200100202001000121212002211101 |
4 | 133311130202320003013113 |
5 | 121322430013321003341 |
6 | 1213440421105450531 |
7 | 41330356301420314 |
oct | 3765344270030727 |
9 | 610661017762741 |
10 | 140012224000471 |
11 | 4068093a378779 |
12 | 13853382640a47 |
13 | 60181330b3784 |
14 | 26808b0977d0b |
15 | 112c08c757831 |
hex | 7f5722e031d7 |
140012224000471 has 2 divisors, whose sum is σ = 140012224000472. Its totient is φ = 140012224000470.
The previous prime is 140012224000423. The next prime is 140012224000507. The reversal of 140012224000471 is 174000422210041.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 140012224000471 - 215 = 140012223967703 is a prime.
It is a super-3 number, since 3×1400122240004713 (a number of 43 digits) contains 333 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (140012224400471) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 70006112000235 + 70006112000236.
It is an arithmetic number, because the mean of its divisors is an integer number (70006112000236).
Almost surely, 2140012224000471 is an apocalyptic number.
140012224000471 is a deficient number, since it is larger than the sum of its proper divisors (1).
140012224000471 is an equidigital number, since it uses as much as digits as its factorization.
140012224000471 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 3584, while the sum is 28.
The spelling of 140012224000471 in words is "one hundred forty trillion, twelve billion, two hundred twenty-four million, four hundred seventy-one", and thus it is an aban number.
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