Base | Representation |
---|---|
bin | 11101010011010000010… |
… | …111000110001111001101 |
3 | 21010111022002111222000111 |
4 | 131103100113012033031 |
5 | 230442212321112232 |
6 | 4141001355000021 |
7 | 265321260044002 |
oct | 35232027061715 |
9 | 7114262458014 |
10 | 2013540410317 |
11 | 706a35083537 |
12 | 2862a2739011 |
13 | 117b50899697 |
14 | 6d654ccd8a9 |
15 | 3759bacb047 |
hex | 1d4d05c63cd |
2013540410317 has 2 divisors, whose sum is σ = 2013540410318. Its totient is φ = 2013540410316.
The previous prime is 2013540410269. The next prime is 2013540410351. The reversal of 2013540410317 is 7130140453102.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 1887727610916 + 125812799401 = 1373946^2 + 354701^2 .
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-2013540410317 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (2013540410017) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1006770205158 + 1006770205159.
It is an arithmetic number, because the mean of its divisors is an integer number (1006770205159).
Almost surely, 22013540410317 is an apocalyptic number.
It is an amenable number.
2013540410317 is a deficient number, since it is larger than the sum of its proper divisors (1).
2013540410317 is an equidigital number, since it uses as much as digits as its factorization.
2013540410317 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 10080, while the sum is 31.
Adding to 2013540410317 its reverse (7130140453102), we get a palindrome (9143680863419).
The spelling of 2013540410317 in words is "two trillion, thirteen billion, five hundred forty million, four hundred ten thousand, three hundred seventeen".
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