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50012142433 = 114546558403
BaseRepresentation
bin101110100100111101…
…001011101101100001
311210002102200001220121
4232210331023231201
51304411102024213
634550353440241
73420230124301
oct564475135541
9153072601817
1050012142433
111a234606920
129838b70081
1349404317b6
1425c61b1601
15147a99288d
hexba4f4bb61

50012142433 has 4 divisors (see below), whose sum is σ = 54558700848. Its totient is φ = 45465584020.

The previous prime is 50012142413. The next prime is 50012142541. The reversal of 50012142433 is 33424121005.

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 a de Polignac number, because none of the positive numbers 2k-50012142433 is a prime.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 50012142398 and 50012142407.

It is not an unprimeable number, because it can be changed into a prime (50012142413) 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, 2273279191 + ... + 2273279212.

It is an arithmetic number, because the mean of its divisors is an integer number (13639675212).

Almost surely, 250012142433 is an apocalyptic number.

It is an amenable number.

50012142433 is a deficient number, since it is larger than the sum of its proper divisors (4546558415).

50012142433 is a wasteful number, since it uses less digits than its factorization.

50012142433 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 4546558414.

The product of its (nonzero) digits is 2880, while the sum is 25.

Adding to 50012142433 its reverse (33424121005), we get a palindrome (83436263438).

The spelling of 50012142433 in words is "fifty billion, twelve million, one hundred forty-two thousand, four hundred thirty-three".

Divisors: 1 11 4546558403 50012142433