Base | Representation |
---|---|
bin | 1100000010110000011011… |
… | …1110111101000110000011 |
3 | 1201212212121222212201121212 |
4 | 3000230012332331012003 |
5 | 3213422043320340120 |
6 | 44055021200240335 |
7 | 2534444514205013 |
oct | 300540676750603 |
9 | 51785558781555 |
10 | 13241501340035 |
11 | 4245764334a21 |
12 | 159a3600100ab |
13 | 75088c0cb017 |
14 | 33ac6bc45c43 |
15 | 17e6963046c5 |
hex | c0b06fbd183 |
13241501340035 has 8 divisors (see below), whose sum is σ = 15892410379056. Its totient is φ = 10591461891360.
The previous prime is 13241501340019. The next prime is 13241501340203. The reversal of 13241501340035 is 53004310514231.
It is a sphenic number, since it is the product of 3 distinct primes.
It is not a de Polignac number, because 13241501340035 - 24 = 13241501340019 is a prime.
It is a Duffinian number.
It is a junction number, because it is equal to n+sod(n) for n = 13241501339983 and 13241501340010.
It is an unprimeable number.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 217364084 + ... + 217424993.
It is an arithmetic number, because the mean of its divisors is an integer number (1986551297382).
Almost surely, 213241501340035 is an apocalyptic number.
13241501340035 is a deficient number, since it is larger than the sum of its proper divisors (2650909039021).
13241501340035 is an equidigital number, since it uses as much as digits as its factorization.
13241501340035 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 434795173.
The product of its (nonzero) digits is 21600, while the sum is 32.
Adding to 13241501340035 its reverse (53004310514231), we get a palindrome (66245811854266).
The spelling of 13241501340035 in words is "thirteen trillion, two hundred forty-one billion, five hundred one million, three hundred forty thousand, thirty-five".
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