Base | Representation |
---|---|
bin | 11101011011101110111… |
… | …00101101110101001100 |
3 | 10120200101201111122000100 |
4 | 32231313130231311030 |
5 | 113032141230043040 |
6 | 2052332233531100 |
7 | 133031325323010 |
oct | 16556734556514 |
9 | 3520351448010 |
10 | 1011321331020 |
11 | 35a998587510 |
12 | 144001573a90 |
13 | 744a094c190 |
14 | 36d3bc80540 |
15 | 1b490629b30 |
hex | eb7772dd4c |
1011321331020 has 576 divisors, whose sum is σ = 4361133643776. Its totient is φ = 182568591360.
The previous prime is 1011321330989. The next prime is 1011321331057. The reversal of 1011321331020 is 201331231101.
1011321331020 is a `hidden beast` number, since 1 + 0 + 11 + 321 + 3 + 310 + 20 = 666.
It is a Harshad number since it is a multiple of its sum of digits (18).
It is a junction number, because it is equal to n+sod(n) for n = 1011321330984 and 1011321331002.
It is an unprimeable number.
It is a polite number, since it can be written in 191 ways as a sum of consecutive naturals, for example, 2897977 + ... + 3228143.
It is an arithmetic number, because the mean of its divisors is an integer number (7571412576).
Almost surely, 21011321331020 is an apocalyptic number.
1011321331020 is a gapful number since it is divisible by the number (10) formed by its first and last digit.
It is an amenable number.
It is a practical number, because each smaller number is the sum of distinct divisors of 1011321331020, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (2180566821888).
1011321331020 is an abundant number, since it is smaller than the sum of its proper divisors (3349812312756).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
1011321331020 is a wasteful number, since it uses less digits than its factorization.
1011321331020 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 330230 (or 330225 counting only the distinct ones).
The product of its (nonzero) digits is 108, while the sum is 18.
Adding to 1011321331020 its reverse (201331231101), we get a palindrome (1212652562121).
It can be divided in two parts, 1011321 and 331020, that added together give a triangular number (1342341 = T1638).
The spelling of 1011321331020 in words is "one trillion, eleven billion, three hundred twenty-one million, three hundred thirty-one thousand, twenty".
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