Base | Representation |
---|---|
bin | 10001101100111001110… |
… | …001000000100010000101 |
3 | 11022021212010200020200022 |
4 | 101230321301000202011 |
5 | 124412234343041232 |
6 | 2330454351424525 |
7 | 153612431663303 |
oct | 21547161004205 |
9 | 4267763606608 |
10 | 1216444893317 |
11 | 42998947a915 |
12 | 177908b25145 |
13 | 8a9305a0404 |
14 | 42c3a5c9273 |
15 | 219987b1812 |
hex | 11b39c40885 |
1216444893317 has 2 divisors, whose sum is σ = 1216444893318. Its totient is φ = 1216444893316.
The previous prime is 1216444893283. The next prime is 1216444893347. The reversal of 1216444893317 is 7133984446121.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 1004968155361 + 211476737956 = 1002481^2 + 459866^2 .
It is a cyclic number.
It is not a de Polignac number, because 1216444893317 - 26 = 1216444893253 is a prime.
It is a super-2 number, since 2×12164448933172 (a number of 25 digits) contains 22 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (1216444893347) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 608222446658 + 608222446659.
It is an arithmetic number, because the mean of its divisors is an integer number (608222446659).
Almost surely, 21216444893317 is an apocalyptic number.
It is an amenable number.
1216444893317 is a deficient number, since it is larger than the sum of its proper divisors (1).
1216444893317 is an equidigital number, since it uses as much as digits as its factorization.
1216444893317 is an evil number, because the sum of its binary digits is even.
The product of its digits is 3483648, while the sum is 53.
The spelling of 1216444893317 in words is "one trillion, two hundred sixteen billion, four hundred forty-four million, eight hundred ninety-three thousand, three hundred seventeen".
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