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2003234113 = 3697354181
BaseRepresentation
bin111011101100110…
…1110110101000001
312011121102211121001
41313121232311001
513100311442423
6530440132001
7100433131456
oct16731566501
95147384531
102003234113
11938856491
1247aa67001
1325c03a43a
1415009b82d
15bad00cad
hex7766ed41

2003234113 has 4 divisors (see below), whose sum is σ = 2003325268. Its totient is φ = 2003142960.

The previous prime is 2003234089. The next prime is 2003234137. The reversal of 2003234113 is 3114323002.

It is a semiprime because it is the product of two primes, and also a brilliant number, because the two primes have the same length.

It is an interprime number because it is at equal distance from previous prime (2003234089) and next prime (2003234137).

It can be written as a sum of positive squares in 2 ways, for example, as 132066064 + 1871168049 = 11492^2 + 43257^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-2003234113 is a prime.

It is a super-2 number, since 2×20032341132 = 8025893822973793538, which contains 22 as substring.

It is a Duffinian number.

It is not an unprimeable number, because it can be changed into a prime (2003234153) 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, 9883 + ... + 64063.

It is an arithmetic number, because the mean of its divisors is an integer number (500831317).

Almost surely, 22003234113 is an apocalyptic number.

It is an amenable number.

2003234113 is a deficient number, since it is larger than the sum of its proper divisors (91155).

2003234113 is an equidigital number, since it uses as much as digits as its factorization.

2003234113 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 91154.

The product of its (nonzero) digits is 432, while the sum is 19.

The square root of 2003234113 is about 44757.5034267998. The cubic root of 2003234113 is about 1260.5998053387.

Adding to 2003234113 its reverse (3114323002), we get a palindrome (5117557115).

Subtracting 2003234113 from its reverse (3114323002), we obtain a square (1111088889 = 333332).

It can be divided in two parts, 20032 and 34113, that added together give a palindrome (54145).

The spelling of 2003234113 in words is "two billion, three million, two hundred thirty-four thousand, one hundred thirteen".

Divisors: 1 36973 54181 2003234113