Base | Representation |
---|---|
bin | 1100110001110101011111… |
… | …1000111000011001110111 |
3 | 1211202012022102212112000021 |
4 | 3030131113320320121313 |
5 | 3320200020003212403 |
6 | 45514342515430011 |
7 | 2650046562120550 |
oct | 314352770703167 |
9 | 54665272775007 |
10 | 14050312554103 |
11 | 4527782167211 |
12 | 16ab0652b5307 |
13 | 7abc28698a17 |
14 | 3680784bd327 |
15 | 195732e318bd |
hex | cc757e38677 |
14050312554103 has 64 divisors (see below), whose sum is σ = 16348558786560. Its totient is φ = 11827892160000.
The previous prime is 14050312554049. The next prime is 14050312554193. The reversal of 14050312554103 is 30145521305041.
It is a de Polignac number, because none of the positive numbers 2k-14050312554103 is a prime.
It is a super-2 number, since 2×140503125541032 (a number of 27 digits) contains 22 as substring.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (14050312554193) by changing a digit.
It is a polite number, since it can be written in 63 ways as a sum of consecutive naturals, for example, 32005267558 + ... + 32005267996.
It is an arithmetic number, because the mean of its divisors is an integer number (255446231040).
Almost surely, 214050312554103 is an apocalyptic number.
14050312554103 is a deficient number, since it is larger than the sum of its proper divisors (2298246232457).
14050312554103 is a wasteful number, since it uses less digits than its factorization.
14050312554103 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 1508.
The product of its (nonzero) digits is 36000, while the sum is 34.
Adding to 14050312554103 its reverse (30145521305041), we get a palindrome (44195833859144).
The spelling of 14050312554103 in words is "fourteen trillion, fifty billion, three hundred twelve million, five hundred fifty-four thousand, one hundred three".
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