Base | Representation |
---|---|
bin | 10001101011011111100… |
… | …011110110000010111011 |
3 | 11022010221122221112210121 |
4 | 101223133203312002323 |
5 | 124401134430234201 |
6 | 2330044250154111 |
7 | 153530064204523 |
oct | 21533743660273 |
9 | 4263848845717 |
10 | 1214931493051 |
11 | 4292821778a1 |
12 | 177566122937 |
13 | 8a74bb989b3 |
14 | 42b355d8683 |
15 | 2190a9bb9a1 |
hex | 11adf8f60bb |
1214931493051 has 2 divisors, whose sum is σ = 1214931493052. Its totient is φ = 1214931493050.
The previous prime is 1214931493019. The next prime is 1214931493187. The reversal of 1214931493051 is 1503941394121.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 1214931493051 - 25 = 1214931493019 is a prime.
It is a super-2 number, since 2×12149314930512 (a number of 25 digits) contains 22 as substring.
It is a junction number, because it is equal to n+sod(n) for n = 1214931492989 and 1214931493007.
It is not a weakly prime, because it can be changed into another prime (1214931493351) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 607465746525 + 607465746526.
It is an arithmetic number, because the mean of its divisors is an integer number (607465746526).
Almost surely, 21214931493051 is an apocalyptic number.
1214931493051 is a deficient number, since it is larger than the sum of its proper divisors (1).
1214931493051 is an equidigital number, since it uses as much as digits as its factorization.
1214931493051 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 116640, while the sum is 43.
The spelling of 1214931493051 in words is "one trillion, two hundred fourteen billion, nine hundred thirty-one million, four hundred ninety-three thousand, fifty-one".
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