Base | Representation |
---|---|
bin | 11101101011100001010011… |
… | …000101111101111110101111 |
3 | 122010011221011200201221201102 |
4 | 131223201103011331332233 |
5 | 114102132203233233201 |
6 | 1141342302111233315 |
7 | 36331526126016035 |
oct | 3553412305757657 |
9 | 563157150657642 |
10 | 130534040133551 |
11 | 386571a27048aa |
12 | 12782455b1883b |
13 | 57ab3cbacc002 |
14 | 2433c4a22a555 |
15 | 101575385726b |
hex | 76b85317dfaf |
130534040133551 has 4 divisors (see below), whose sum is σ = 130725158787432. Its totient is φ = 130342921479672.
The previous prime is 130534040133529. The next prime is 130534040133559. The reversal of 130534040133551 is 155331040435031.
It is a semiprime because it is the product of two primes.
It is a cyclic number.
It is not a de Polignac number, because 130534040133551 - 238 = 130259162226607 is a prime.
It is a Duffinian number.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (130534040133559) 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, 95559325916 + ... + 95559327281.
It is an arithmetic number, because the mean of its divisors is an integer number (32681289696858).
Almost surely, 2130534040133551 is an apocalyptic number.
130534040133551 is a deficient number, since it is larger than the sum of its proper divisors (191118653881).
130534040133551 is an equidigital number, since it uses as much as digits as its factorization.
130534040133551 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 191118653880.
The product of its (nonzero) digits is 162000, while the sum is 38.
Adding to 130534040133551 its reverse (155331040435031), we get a palindrome (285865080568582).
The spelling of 130534040133551 in words is "one hundred thirty trillion, five hundred thirty-four billion, forty million, one hundred thirty-three thousand, five hundred fifty-one".
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