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113031444403 = 472404924349
BaseRepresentation
bin110100101000100110…
…0101111011110110011
3101210202100121022022201
41221101030233132303
53322442022210103
6123532000044031
711111011265533
oct1512114573663
9353670538281
10113031444403
1143a33338169
1219aa6003617
13a8745709a7
145683a534c3
152e182ed51d
hex1a5132f7b3

113031444403 has 4 divisors (see below), whose sum is σ = 115436368800. Its totient is φ = 110626520008.

The previous prime is 113031444349. The next prime is 113031444419. The reversal of 113031444403 is 304444130311.

It is a happy number.

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 113031444403 - 213 = 113031436211 is a prime.

It is a super-2 number, since 2×1130314444032 (a number of 23 digits) contains 22 as substring.

It is a Duffinian number.

It is not an unprimeable number, because it can be changed into a prime (113031444433) by changing a digit.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 1202462128 + ... + 1202462221.

It is an arithmetic number, because the mean of its divisors is an integer number (28859092200).

Almost surely, 2113031444403 is an apocalyptic number.

113031444403 is a deficient number, since it is larger than the sum of its proper divisors (2404924397).

113031444403 is an equidigital number, since it uses as much as digits as its factorization.

113031444403 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 2404924396.

The product of its (nonzero) digits is 6912, while the sum is 28.

Adding to 113031444403 its reverse (304444130311), we get a palindrome (417475574714).

The spelling of 113031444403 in words is "one hundred thirteen billion, thirty-one million, four hundred forty-four thousand, four hundred three".

Divisors: 1 47 2404924349 113031444403