Base | Representation |
---|---|
bin | 1100110101011101001111… |
… | …0000011000000110010001 |
3 | 1211222010212210121110121100 |
4 | 3031113103300120012101 |
5 | 3322204420140212203 |
6 | 50003111410215013 |
7 | 2654411256543036 |
oct | 315272360300621 |
9 | 54863783543540 |
10 | 14112520241553 |
11 | 45510a4844558 |
12 | 16bb126584469 |
13 | 7b4a60619636 |
14 | 36b09a25138d |
15 | 1971743bdea3 |
hex | cd5d3c18191 |
14112520241553 has 24 divisors (see below), whose sum is σ = 20458164066912. Its totient is φ = 9374468580288.
The previous prime is 14112520241527. The next prime is 14112520241623. The reversal of 14112520241553 is 35514202521141.
14112520241553 is a `hidden beast` number, since 1 + 4 + 1 + 12 + 52 + 0 + 2 + 41 + 553 = 666.
It is not a de Polignac number, because 14112520241553 - 26 = 14112520241489 is a prime.
It is not an unprimeable number, because it can be changed into a prime (14112520241153) by changing a digit.
It is a polite number, since it can be written in 23 ways as a sum of consecutive naturals, for example, 39153925 + ... + 39512717.
It is an arithmetic number, because the mean of its divisors is an integer number (852423502788).
Almost surely, 214112520241553 is an apocalyptic number.
It is an amenable number.
14112520241553 is a deficient number, since it is larger than the sum of its proper divisors (6345643825359).
14112520241553 is a wasteful number, since it uses less digits than its factorization.
14112520241553 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 374525 (or 374522 counting only the distinct ones).
The product of its (nonzero) digits is 48000, while the sum is 36.
Adding to 14112520241553 its reverse (35514202521141), we get a palindrome (49626722762694).
The spelling of 14112520241553 in words is "fourteen trillion, one hundred twelve billion, five hundred twenty million, two hundred forty-one thousand, five hundred fifty-three".
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