Base | Representation |
---|---|
bin | 1100110101011101001110… |
… | …1111100000100110101101 |
3 | 1211222010212210011220210211 |
4 | 3031113103233200212231 |
5 | 3322204420110424003 |
6 | 50003111401302421 |
7 | 2654411254605253 |
oct | 315272357404655 |
9 | 54863783156724 |
10 | 14112520014253 |
11 | 45510a46a9801 |
12 | 16bb126494a11 |
13 | 7b4a6056a03b |
14 | 36b09a1d25d3 |
15 | 19717437696d |
hex | cd5d3be09ad |
14112520014253 has 8 divisors (see below), whose sum is σ = 14171802772160. Its totient is φ = 14053243245120.
The previous prime is 14112520014239. The next prime is 14112520014269. The reversal of 14112520014253 is 35241002521141.
It is a sphenic number, since it is the product of 3 distinct primes.
It is a cyclic number.
It is not a de Polignac number, because 14112520014253 - 25 = 14112520014221 is a prime.
It is a Duffinian number.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (14112520014223) by changing a digit.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 3257338 + ... + 6231796.
It is an arithmetic number, because the mean of its divisors is an integer number (1771475346520).
Almost surely, 214112520014253 is an apocalyptic number.
It is an amenable number.
14112520014253 is a deficient number, since it is larger than the sum of its proper divisors (59282757907).
14112520014253 is a wasteful number, since it uses less digits than its factorization.
14112520014253 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 2994387.
The product of its (nonzero) digits is 9600, while the sum is 31.
Adding to 14112520014253 its reverse (35241002521141), we get a palindrome (49353522535394).
The spelling of 14112520014253 in words is "fourteen trillion, one hundred twelve billion, five hundred twenty million, fourteen thousand, two hundred fifty-three".
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