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2551015113 = 3850338371
BaseRepresentation
bin1001100000001101…
…0110011011001001
320120210011222212020
42120003112123021
520211024440423
61101045035053
7120134161563
oct23003263311
96523158766
102551015113
111099a88459
125b23b5a89
133186818ca
141a2b30333
15ede564e3
hex980d66c9

2551015113 has 4 divisors (see below), whose sum is σ = 3401353488. Its totient is φ = 1700676740.

The previous prime is 2551015063. The next prime is 2551015127. The reversal of 2551015113 is 3115101552.

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 a cyclic number.

It is not a de Polignac number, because 2551015113 - 228 = 2282579657 is a prime.

It is not an unprimeable number, because it can be changed into a prime (2551016113) 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, 425169183 + ... + 425169188.

It is an arithmetic number, because the mean of its divisors is an integer number (850338372).

Almost surely, 22551015113 is an apocalyptic number.

It is an amenable number.

2551015113 is a deficient number, since it is larger than the sum of its proper divisors (850338375).

2551015113 is an equidigital number, since it uses as much as digits as its factorization.

2551015113 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 850338374.

The product of its (nonzero) digits is 750, while the sum is 24.

The square root of 2551015113 is about 50507.5748081414. The cubic root of 2551015113 is about 1366.3784708656.

Adding to 2551015113 its reverse (3115101552), we get a palindrome (5666116665).

The spelling of 2551015113 in words is "two billion, five hundred fifty-one million, fifteen thousand, one hundred thirteen".

Divisors: 1 3 850338371 2551015113