Base | Representation |
---|---|
bin | 1100010010000000100110… |
… | …1101001001110101100101 |
3 | 1202210220222220212101210010 |
4 | 3010200021231021311211 |
5 | 3232220222220312331 |
6 | 44415235102320433 |
7 | 2562412211150643 |
oct | 304401155116545 |
9 | 52726886771703 |
10 | 13503540010341 |
11 | 4336901332377 |
12 | 16210b0289119 |
13 | 76c4bc340c64 |
14 | 349809460193 |
15 | 1863d0e7be46 |
hex | c4809b49d65 |
13503540010341 has 16 divisors (see below), whose sum is σ = 18258579947520. Its totient is φ = 8875433822640.
The previous prime is 13503540010333. The next prime is 13503540010369. The reversal of 13503540010341 is 14301004530531.
It is not a de Polignac number, because 13503540010341 - 23 = 13503540010333 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 13503540010299 and 13503540010308.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (13503540010321) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (19) of ones.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 15032220 + ... + 15905178.
It is an arithmetic number, because the mean of its divisors is an integer number (1141161246720).
Almost surely, 213503540010341 is an apocalyptic number.
It is an amenable number.
13503540010341 is a deficient number, since it is larger than the sum of its proper divisors (4755039937179).
13503540010341 is an equidigital number, since it uses as much as digits as its factorization.
13503540010341 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 945656.
The product of its (nonzero) digits is 10800, while the sum is 30.
Adding to 13503540010341 its reverse (14301004530531), we get a palindrome (27804544540872).
The spelling of 13503540010341 in words is "thirteen trillion, five hundred three billion, five hundred forty million, ten thousand, three hundred forty-one".
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