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13025345035 = 53796873533
BaseRepresentation
bin11000010000101111…
…01111111000001011
31020121202111002022111
430020113233320023
5203133442020120
65552254203151
7640531414603
oct141027577013
936552432274
1013025345035
11558451656a
1226361b34b7
1312c7706c71
148b7c85c03
155137ab75a
hex3085efe0b

13025345035 has 8 divisors (see below), whose sum is σ = 15671657520. Its totient is φ = 10392780384.

The previous prime is 13025345009. The next prime is 13025345059. The reversal of 13025345035 is 53054352031.

It is a sphenic number, since it is the product of 3 distinct primes.

It is a cyclic number.

It is not a de Polignac number, because 13025345035 - 211 = 13025342987 is a prime.

It is a super-2 number, since 2×130253450352 (a number of 21 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 13025344988 and 13025345006.

It is an unprimeable number.

It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 3434872 + ... + 3438661.

It is an arithmetic number, because the mean of its divisors is an integer number (1958957190).

Almost surely, 213025345035 is an apocalyptic number.

13025345035 is a deficient number, since it is larger than the sum of its proper divisors (2646312485).

13025345035 is an equidigital number, since it uses as much as digits as its factorization.

13025345035 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 6873917.

The product of its (nonzero) digits is 27000, while the sum is 31.

Adding to 13025345035 its reverse (53054352031), we get a palindrome (66079697066).

The spelling of 13025345035 in words is "thirteen billion, twenty-five million, three hundred forty-five thousand, thirty-five".

Divisors: 1 5 379 1895 6873533 34367665 2605069007 13025345035