Base | Representation |
---|---|
bin | 1000110111111001110011… |
… | …00000000011101111111101 |
3 | 2120002102110121011200022210 |
4 | 10123330321200003233331 |
5 | 10024200111242143010 |
6 | 105300052102131033 |
7 | 4052524233650244 |
oct | 433747140035775 |
9 | 76072417150283 |
10 | 19513001131005 |
11 | 6243465487808 |
12 | 2231900720a79 |
13 | ab70b68426b5 |
14 | 4b66119b995b |
15 | 23c8a0829b20 |
hex | 11bf39803bfd |
19513001131005 has 8 divisors (see below), whose sum is σ = 31220801809632. Its totient is φ = 10406933936528.
The previous prime is 19513001130949. The next prime is 19513001131021. The reversal of 19513001131005 is 50013110031591.
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 19513001131005 - 26 = 19513001130941 is a prime.
It is a congruent number.
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, 650433371019 + ... + 650433371048.
It is an arithmetic number, because the mean of its divisors is an integer number (3902600226204).
Almost surely, 219513001131005 is an apocalyptic number.
19513001131005 is a gapful number since it is divisible by the number (15) formed by its first and last digit.
It is an amenable number.
19513001131005 is a deficient number, since it is larger than the sum of its proper divisors (11707800678627).
19513001131005 is a wasteful number, since it uses less digits than its factorization.
19513001131005 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 1300866742075.
The product of its (nonzero) digits is 2025, while the sum is 30.
Adding to 19513001131005 its reverse (50013110031591), we get a palindrome (69526111162596).
The spelling of 19513001131005 in words is "nineteen trillion, five hundred thirteen billion, one million, one hundred thirty-one thousand, five".
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