Base | Representation |
---|---|
bin | 11100110100010101001… |
… | …111111010000011011111 |
3 | 21000022121001100020021001 |
4 | 130310111033322003133 |
5 | 224421212110234233 |
6 | 4113430510051131 |
7 | 263034403534525 |
oct | 34642517720337 |
9 | 7008531306231 |
10 | 1980336414943 |
11 | 6a3946308347 |
12 | 27b9767a9aa7 |
13 | 11498b797343 |
14 | 6bbc513a315 |
15 | 367a6a6527d |
hex | 1cd153fa0df |
1980336414943 has 2 divisors, whose sum is σ = 1980336414944. Its totient is φ = 1980336414942.
The previous prime is 1980336414941. The next prime is 1980336414949. The reversal of 1980336414943 is 3494146330891.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 1980336414943 - 21 = 1980336414941 is a prime.
Together with 1980336414941, it forms a pair of twin primes.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (1980336414941) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 990168207471 + 990168207472.
It is an arithmetic number, because the mean of its divisors is an integer number (990168207472).
Almost surely, 21980336414943 is an apocalyptic number.
1980336414943 is a deficient number, since it is larger than the sum of its proper divisors (1).
1980336414943 is an equidigital number, since it uses as much as digits as its factorization.
1980336414943 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 6718464, while the sum is 55.
The spelling of 1980336414943 in words is "one trillion, nine hundred eighty billion, three hundred thirty-six million, four hundred fourteen thousand, nine hundred forty-three".
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