Base | Representation |
---|---|
bin | 1100110101011011101101… |
… | …1100101110000100101101 |
3 | 1211222002211101020202212111 |
4 | 3031112323130232010231 |
5 | 3322203101240414001 |
6 | 50003003121404021 |
7 | 2654365205340103 |
oct | 315267334560455 |
9 | 54862741222774 |
10 | 14112112435501 |
11 | 4550a05627863 |
12 | 16bb031a99611 |
13 | 7b49c6ca3c92 |
14 | 36b05c017c73 |
15 | 19714d6b7851 |
hex | cd5bb72e12d |
14112112435501 has 2 divisors, whose sum is σ = 14112112435502. Its totient is φ = 14112112435500.
The previous prime is 14112112435483. The next prime is 14112112435523. The reversal of 14112112435501 is 10553421121141.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 11739360375625 + 2372752059876 = 3426275^2 + 1540374^2 .
It is a cyclic number.
It is not a de Polignac number, because 14112112435501 - 221 = 14112110338349 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (14112112475501) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 7056056217750 + 7056056217751.
It is an arithmetic number, because the mean of its divisors is an integer number (7056056217751).
Almost surely, 214112112435501 is an apocalyptic number.
It is an amenable number.
14112112435501 is a deficient number, since it is larger than the sum of its proper divisors (1).
14112112435501 is an equidigital number, since it uses as much as digits as its factorization.
14112112435501 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 4800, while the sum is 31.
Adding to 14112112435501 its reverse (10553421121141), we get a palindrome (24665533556642).
The spelling of 14112112435501 in words is "fourteen trillion, one hundred twelve billion, one hundred twelve million, four hundred thirty-five thousand, five hundred one".
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