Search a number
-
+
13300531404214 = 26650265702107
BaseRepresentation
bin1100000110001100010101…
…1100110001110110110110
31202002111222220100200120011
43001203011130301312312
53220403442014413324
644142102455330434
72541634365365335
oct301430534616666
952074886320504
1013300531404214
1142687a626292a
1215a9895080a1a
13756308916292
1433da6b975d1c
15180e9d813c94
hexc18c5731db6

13300531404214 has 4 divisors (see below), whose sum is σ = 19950797106324. Its totient is φ = 6650265702106.

The previous prime is 13300531404211. The next prime is 13300531404227. The reversal of 13300531404214 is 41240413500331.

13300531404214 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a semiprime because it is the product of two primes.

It is a congruent number.

It is not an unprimeable number, because it can be changed into a prime (13300531404211) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 3325132851052 + ... + 3325132851055.

It is an arithmetic number, because the mean of its divisors is an integer number (4987699276581).

Almost surely, 213300531404214 is an apocalyptic number.

13300531404214 is a deficient number, since it is larger than the sum of its proper divisors (6650265702110).

13300531404214 is an equidigital number, since it uses as much as digits as its factorization.

13300531404214 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 6650265702109.

The product of its (nonzero) digits is 17280, while the sum is 31.

Adding to 13300531404214 its reverse (41240413500331), we get a palindrome (54540944904545).

The spelling of 13300531404214 in words is "thirteen trillion, three hundred billion, five hundred thirty-one million, four hundred four thousand, two hundred fourteen".

Divisors: 1 2 6650265702107 13300531404214