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105113433 = 335037811
BaseRepresentation
bin1100100001111…
…10011101011001
321022210022112010
412100332131121
5203402112213
614232540133
72414310105
oct620763531
9238708463
10105113433
1154374307
122b251649
1318a140c9
14dd62905
159364ac3
hex643e759

105113433 has 4 divisors (see below), whose sum is σ = 140151248. Its totient is φ = 70075620.

The previous prime is 105113399. The next prime is 105113447. The reversal of 105113433 is 334311501.

It is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4.

It is not a de Polignac number, because 105113433 - 26 = 105113369 is a prime.

It is a super-2 number, since 2×1051134332 = 22097667594090978, which contains 22 as substring.

It is a Duffinian number.

It is not an unprimeable number, because it can be changed into a prime (105113233) 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, 17518903 + ... + 17518908.

It is an arithmetic number, because the mean of its divisors is an integer number (35037812).

Almost surely, 2105113433 is an apocalyptic number.

It is an amenable number.

105113433 is a deficient number, since it is larger than the sum of its proper divisors (35037815).

105113433 is an equidigital number, since it uses as much as digits as its factorization.

105113433 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 35037814.

The product of its (nonzero) digits is 540, while the sum is 21.

The square root of 105113433 is about 10252.4842355402. The cubic root of 105113433 is about 471.9392232982.

Adding to 105113433 its reverse (334311501), we get a palindrome (439424934).

The spelling of 105113433 in words is "one hundred five million, one hundred thirteen thousand, four hundred thirty-three".

Divisors: 1 3 35037811 105113433