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51125155472841 = 317041718490947
BaseRepresentation
bin10111001111111100000001…
…01110110010010111001001
320201000111221222200110022220
423213332000232302113021
523200113304300112331
6300422323332004253
713524446004222462
oct1347760056622711
9221014858613286
1051125155472841
111532106948615a
125898493160689
13226b109ccb708
14c8a683969d69
155d9d38c94296
hex2e7f80bb25c9

51125155472841 has 4 divisors (see below), whose sum is σ = 68166873963792. Its totient is φ = 34083436981892.

The previous prime is 51125155472821. The next prime is 51125155472891. The reversal of 51125155472841 is 14827455152115.

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 51125155472841 - 217 = 51125155341769 is a prime.

It is a super-2 number, since 2×511251554728412 (a number of 28 digits) contains 22 as substring.

It is not an unprimeable number, because it can be changed into a prime (51125155472801) 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, 8520859245471 + ... + 8520859245476.

It is an arithmetic number, because the mean of its divisors is an integer number (17041718490948).

Almost surely, 251125155472841 is an apocalyptic number.

It is an amenable number.

51125155472841 is a deficient number, since it is larger than the sum of its proper divisors (17041718490951).

51125155472841 is a wasteful number, since it uses less digits than its factorization.

51125155472841 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 17041718490950.

The product of its digits is 2240000, while the sum is 51.

The spelling of 51125155472841 in words is "fifty-one trillion, one hundred twenty-five billion, one hundred fifty-five million, four hundred seventy-two thousand, eight hundred forty-one".

Divisors: 1 3 17041718490947 51125155472841