Base | Representation |
---|---|
bin | 1001101001000010110001… |
… | …00010011011010101001011 |
3 | 2210001211122002022001111222 |
4 | 10310201120202123111023 |
5 | 10234331024130233112 |
6 | 113031450420312255 |
7 | 4315516355325533 |
oct | 464413042332513 |
9 | 83054562261458 |
10 | 21201443992907 |
11 | 68345334603a2 |
12 | 2464b9484268b |
13 | baa397ab12c9 |
14 | 53422646a4c3 |
15 | 26b771706372 |
hex | 13485889b54b |
21201443992907 has 2 divisors, whose sum is σ = 21201443992908. Its totient is φ = 21201443992906.
The previous prime is 21201443992853. The next prime is 21201443992931. The reversal of 21201443992907 is 70929934410212.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 21201443992907 - 236 = 21132724516171 is a prime.
It is a super-3 number, since 3×212014439929073 (a number of 41 digits) contains 333 as substring.
It is not a weakly prime, because it can be changed into another prime (21201443992987) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 10600721996453 + 10600721996454.
It is an arithmetic number, because the mean of its divisors is an integer number (10600721996454).
Almost surely, 221201443992907 is an apocalyptic number.
21201443992907 is a deficient number, since it is larger than the sum of its proper divisors (1).
21201443992907 is an equidigital number, since it uses as much as digits as its factorization.
21201443992907 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 1959552, while the sum is 53.
The spelling of 21201443992907 in words is "twenty-one trillion, two hundred one billion, four hundred forty-three million, nine hundred ninety-two thousand, nine hundred seven".
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