Base | Representation |
---|---|
bin | 100101001100100001101… |
… | …0011000001001111011011 |
3 | 200002201022211120120001102 |
4 | 1022121003103001033123 |
5 | 1132224130334012111 |
6 | 14512252412542015 |
7 | 1035224412144233 |
oct | 112310323011733 |
9 | 20081284516042 |
10 | 5112140141531 |
11 | 16a1056922685 |
12 | 6a692548190b |
13 | 2b10c3a950b7 |
14 | 139600ca4cc3 |
15 | 8cea266013b |
hex | 4a6434c13db |
5112140141531 has 16 divisors (see below), whose sum is σ = 5555170790400. Its totient is φ = 4686528132000.
The previous prime is 5112140141513. The next prime is 5112140141561. The reversal of 5112140141531 is 1351410412115.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-5112140141531 is a prime.
It is a super-2 number, since 2×51121401415312 (a number of 26 digits) contains 22 as substring.
It is a Duffinian number.
It is a junction number, because it is equal to n+sod(n) for n = 5112140141494 and 5112140141503.
It is not an unprimeable number, because it can be changed into a prime (5112140141561) by changing a digit.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 8455025 + ... + 9039453.
It is an arithmetic number, because the mean of its divisors is an integer number (347198174400).
Almost surely, 25112140141531 is an apocalyptic number.
5112140141531 is a deficient number, since it is larger than the sum of its proper divisors (443030648869).
5112140141531 is a wasteful number, since it uses less digits than its factorization.
5112140141531 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 599330.
The product of its (nonzero) digits is 2400, while the sum is 29.
Adding to 5112140141531 its reverse (1351410412115), we get a palindrome (6463550553646).
The spelling of 5112140141531 in words is "five trillion, one hundred twelve billion, one hundred forty million, one hundred forty-one thousand, five hundred thirty-one".
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