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215200342003 = 330376513919
BaseRepresentation
bin1100100001101011110…
…0001001111111110011
3202120110200102001202011
43020122330021333303
512011212241421003
6242510042103351
721355602503623
oct3103274117763
9676420361664
10215200342003
11832a1a8a119
123585a1aab57
13173a751209a
14a5b6b76883
1558e7b5e46d
hex321af09ff3

215200342003 has 4 divisors (see below), whose sum is σ = 215206888960. Its totient is φ = 215193795048.

The previous prime is 215200341971. The next prime is 215200342037. The reversal of 215200342003 is 300243002512.

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 215200342003 - 25 = 215200341971 is a prime.

It is a super-2 number, since 2×2152003420032 (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 (215200342073) 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, 3223923 + ... + 3289996.

It is an arithmetic number, because the mean of its divisors is an integer number (53801722240).

Almost surely, 2215200342003 is an apocalyptic number.

215200342003 is a deficient number, since it is larger than the sum of its proper divisors (6546957).

215200342003 is an equidigital number, since it uses as much as digits as its factorization.

215200342003 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 6546956.

The product of its (nonzero) digits is 1440, while the sum is 22.

Adding to 215200342003 its reverse (300243002512), we get a palindrome (515443344515).

The spelling of 215200342003 in words is "two hundred fifteen billion, two hundred million, three hundred forty-two thousand, three".

Divisors: 1 33037 6513919 215200342003