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51102543145 = 510220508629
BaseRepresentation
bin101111100101111100…
…101110110100101001
311212220102110011111011
4233211330232310221
51314124222340040
635250504503521
73456240310153
oct574574566451
9155812404434
1051102543145
111a744065035
129aa217aba1
134a85301114
14268ad31ad3
1514e157e3ea
hexbe5f2ed29

51102543145 has 4 divisors (see below), whose sum is σ = 61323051780. Its totient is φ = 40882034512.

The previous prime is 51102543113. The next prime is 51102543161. The reversal of 51102543145 is 54134520115.

It is a semiprime because it is the product of two primes.

It can be written as a sum of positive squares in 2 ways, for example, as 33104710809 + 17997832336 = 181947^2 + 134156^2 .

It is a cyclic number.

It is not a de Polignac number, because 51102543145 - 25 = 51102543113 is a prime.

It is a super-2 number, since 2×511025431452 (a number of 22 digits) contains 22 as substring.

It is an unprimeable number.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 5110254310 + ... + 5110254319.

It is an arithmetic number, because the mean of its divisors is an integer number (15330762945).

Almost surely, 251102543145 is an apocalyptic number.

It is an amenable number.

51102543145 is a deficient number, since it is larger than the sum of its proper divisors (10220508635).

51102543145 is a wasteful number, since it uses less digits than its factorization.

51102543145 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 10220508634.

The product of its (nonzero) digits is 12000, while the sum is 31.

The spelling of 51102543145 in words is "fifty-one billion, one hundred two million, five hundred forty-three thousand, one hundred forty-five".

Divisors: 1 5 10220508629 51102543145