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2003440130114 = 21001720065057
BaseRepresentation
bin11101001001110110010…
…101100101100001000010
321002112020012001120022112
4131021312111211201002
5230311021133130424
64132211230535322
7264513054061643
oct35116625454102
97075205046275
102003440130114
117027217a6525
12284344067542
13116c0119b017
146cd777056ca
15371aa01060e
hex1d276565842

2003440130114 has 4 divisors (see below), whose sum is σ = 3005160195174. Its totient is φ = 1001720065056.

The previous prime is 2003440130071. The next prime is 2003440130117. The reversal of 2003440130114 is 4110310443002.

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

It can be written as a sum of positive squares in only one way, i.e., 1112955371089 + 890484759025 = 1054967^2 + 943655^2 .

It is a self number, because there is not a number n which added to its sum of digits gives 2003440130114.

It is not an unprimeable number, because it can be changed into a prime (2003440130117) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (19) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 500860032527 + ... + 500860032530.

Almost surely, 22003440130114 is an apocalyptic number.

2003440130114 is a deficient number, since it is larger than the sum of its proper divisors (1001720065060).

2003440130114 is a wasteful number, since it uses less digits than its factorization.

2003440130114 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 1001720065059.

The product of its (nonzero) digits is 1152, while the sum is 23.

Adding to 2003440130114 its reverse (4110310443002), we get a palindrome (6113750573116).

The spelling of 2003440130114 in words is "two trillion, three billion, four hundred forty million, one hundred thirty thousand, one hundred fourteen".

Divisors: 1 2 1001720065057 2003440130114