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14910204035 = 52982040807
BaseRepresentation
bin11011110001011011…
…11010100010000011
31102111010011211210002
431320231322202003
5221014003012120
610503305230215
71035326433416
oct157055724203
942433154702
1014910204035
11636146aa81
122a8149236b
131538068c4c
14a1632bb7d
155c3ec8375
hex378b7a883

14910204035 has 4 divisors (see below), whose sum is σ = 17892244848. Its totient is φ = 11928163224.

The previous prime is 14910204017. The next prime is 14910204041. The reversal of 14910204035 is 53040201941.

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

It is a cyclic number.

It is not a de Polignac number, because 14910204035 - 28 = 14910203779 is a prime.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 14910203989 and 14910204007.

It is an unprimeable number.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 1491020399 + ... + 1491020408.

It is an arithmetic number, because the mean of its divisors is an integer number (4473061212).

Almost surely, 214910204035 is an apocalyptic number.

14910204035 is a deficient number, since it is larger than the sum of its proper divisors (2982040813).

14910204035 is an equidigital number, since it uses as much as digits as its factorization.

14910204035 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 2982040812.

The product of its (nonzero) digits is 4320, while the sum is 29.

Adding to 14910204035 its reverse (53040201941), we get a palindrome (67950405976).

The spelling of 14910204035 in words is "fourteen billion, nine hundred ten million, two hundred four thousand, thirty-five".

Divisors: 1 5 2982040807 14910204035