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13002551042155 = 52600510208431
BaseRepresentation
bin1011110100110110010001…
…1011111010000001101011
31201001000212010201220020021
42331031210123322001223
53201013211031322110
643353142310141311
72511255223165153
oct275154433720153
951030763656207
1013002551042155
1141633950a64a4
12155bb94051837
1373419b34940c
1432d480b61763
1517835d5e62da
hexbd3646fa06b

13002551042155 has 4 divisors (see below), whose sum is σ = 15603061250592. Its totient is φ = 10402040833720.

The previous prime is 13002551042147. The next prime is 13002551042171. The reversal of 13002551042155 is 55124015520031.

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

It is not a de Polignac number, because 13002551042155 - 23 = 13002551042147 is a prime.

It is a Duffinian number.

It is a self number, because there is not a number n which added to its sum of digits gives 13002551042155.

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, 1300255104211 + ... + 1300255104220.

It is an arithmetic number, because the mean of its divisors is an integer number (3900765312648).

Almost surely, 213002551042155 is an apocalyptic number.

13002551042155 is a deficient number, since it is larger than the sum of its proper divisors (2600510208437).

13002551042155 is an equidigital number, since it uses as much as digits as its factorization.

13002551042155 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 2600510208436.

The product of its (nonzero) digits is 30000, while the sum is 34.

Adding to 13002551042155 its reverse (55124015520031), we get a palindrome (68126566562186).

The spelling of 13002551042155 in words is "thirteen trillion, two billion, five hundred fifty-one million, forty-two thousand, one hundred fifty-five".

Divisors: 1 5 2600510208431 13002551042155