Base | Representation |
---|---|
bin | 1100001001010011100010… |
… | …0100100001010101111101 |
3 | 1202021122000101111210020121 |
4 | 3002110320210201111331 |
5 | 3222242444104433432 |
6 | 44222424413014541 |
7 | 2545536425466031 |
oct | 302247044412575 |
9 | 52248011453217 |
10 | 13354002421117 |
11 | 4289445257296 |
12 | 15b8118484451 |
13 | 75b37a815457 |
14 | 3424a127c1c1 |
15 | 18257cc8d197 |
hex | c253892157d |
13354002421117 has 2 divisors, whose sum is σ = 13354002421118. Its totient is φ = 13354002421116.
The previous prime is 13354002421079. The next prime is 13354002421121. The reversal of 13354002421117 is 71112420045331.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 13353974267481 + 28153636 = 3654309^2 + 5306^2 .
It is a cyclic number.
It is not a de Polignac number, because 13354002421117 - 27 = 13354002420989 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (13354002421157) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6677001210558 + 6677001210559.
It is an arithmetic number, because the mean of its divisors is an integer number (6677001210559).
Almost surely, 213354002421117 is an apocalyptic number.
It is an amenable number.
13354002421117 is a deficient number, since it is larger than the sum of its proper divisors (1).
13354002421117 is an equidigital number, since it uses as much as digits as its factorization.
13354002421117 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 20160, while the sum is 34.
Adding to 13354002421117 its reverse (71112420045331), we get a palindrome (84466422466448).
The spelling of 13354002421117 in words is "thirteen trillion, three hundred fifty-four billion, two million, four hundred twenty-one thousand, one hundred seventeen".
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