Base | Representation |
---|---|
bin | 1101000000101010010110… |
… | …1001011111010101001001 |
3 | 1212122112202110021012110220 |
4 | 3100022211221133111021 |
5 | 3333333140133300203 |
6 | 50231345140500253 |
7 | 3004334510363220 |
oct | 320124551372511 |
9 | 55575673235426 |
10 | 14305020212553 |
11 | 46157a7a84984 |
12 | 17304aa3aa689 |
13 | 7c9c5ab6a515 |
14 | 37651c213cb7 |
15 | 19c18e1a7653 |
hex | d02a5a5f549 |
14305020212553 has 32 divisors (see below), whose sum is σ = 23201208483840. Its totient is φ = 7653176336640.
The previous prime is 14305020212531. The next prime is 14305020212599. The reversal of 14305020212553 is 35521202050341.
It is not a de Polignac number, because 14305020212553 - 28 = 14305020212297 is a prime.
It is a super-3 number, since 3×143050202125533 (a number of 40 digits) contains 333 as substring.
It is not an unprimeable number, because it can be changed into a prime (14305020213553) by changing a digit.
It is a polite number, since it can be written in 31 ways as a sum of consecutive naturals, for example, 104827323 + ... + 104963696.
It is an arithmetic number, because the mean of its divisors is an integer number (725037765120).
Almost surely, 214305020212553 is an apocalyptic number.
It is an amenable number.
14305020212553 is a deficient number, since it is larger than the sum of its proper divisors (8896188271287).
14305020212553 is a wasteful number, since it uses less digits than its factorization.
14305020212553 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 209791237.
The product of its (nonzero) digits is 36000, while the sum is 33.
Adding to 14305020212553 its reverse (35521202050341), we get a palindrome (49826222262894).
The spelling of 14305020212553 in words is "fourteen trillion, three hundred five billion, twenty million, two hundred twelve thousand, five hundred fifty-three".
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