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2140121433 = 3713373811
BaseRepresentation
bin111111110001111…
…1010100101011001
312112011000102200210
41333203322211121
513340332341213
6552210121333
7104014506042
oct17743724531
95464012623
102140121433
119a90519aa
124b8880249
132814c7a58
141643318c9
15c7d400c3
hex7f8fa959

2140121433 has 4 divisors (see below), whose sum is σ = 2853495248. Its totient is φ = 1426747620.

The previous prime is 2140121407. The next prime is 2140121447. The reversal of 2140121433 is 3341210412.

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 2140121433 - 29 = 2140120921 is a prime.

It is a Duffinian number.

It is not an unprimeable number, because it can be changed into a prime (2140121833) 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, 356686903 + ... + 356686908.

It is an arithmetic number, because the mean of its divisors is an integer number (713373812).

Almost surely, 22140121433 is an apocalyptic number.

It is an amenable number.

2140121433 is a deficient number, since it is larger than the sum of its proper divisors (713373815).

2140121433 is an equidigital number, since it uses as much as digits as its factorization.

2140121433 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 713373814.

The product of its (nonzero) digits is 576, while the sum is 21.

The square root of 2140121433 is about 46261.4465078644. The cubic root of 2140121433 is about 1288.6831168002.

Adding to 2140121433 its reverse (3341210412), we get a palindrome (5481331845).

The spelling of 2140121433 in words is "two billion, one hundred forty million, one hundred twenty-one thousand, four hundred thirty-three".

Divisors: 1 3 713373811 2140121433