Base | Representation |
---|---|
bin | 11011010101100110010101… |
… | …110101110000010001011111 |
3 | 120202201000121021202002122111 |
4 | 123111212111311300101133 |
5 | 111224333140023212341 |
6 | 1103413334241254451 |
7 | 34216304323551214 |
oct | 3325462565602137 |
9 | 522630537662574 |
10 | 120231533413471 |
11 | 353449a82006a7 |
12 | 11599815334427 |
13 | 5211a28701a8c |
14 | 21993470b9b0b |
15 | dd777114d681 |
hex | 6d5995d7045f |
120231533413471 has 2 divisors, whose sum is σ = 120231533413472. Its totient is φ = 120231533413470.
The previous prime is 120231533413427. The next prime is 120231533413487. The reversal of 120231533413471 is 174314335132021.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 120231533413471 - 243 = 111435440391263 is a prime.
It is a super-3 number, since 3×1202315334134713 (a number of 43 digits) contains 333 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (120231533416471) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 60115766706735 + 60115766706736.
It is an arithmetic number, because the mean of its divisors is an integer number (60115766706736).
Almost surely, 2120231533413471 is an apocalyptic number.
120231533413471 is a deficient number, since it is larger than the sum of its proper divisors (1).
120231533413471 is an equidigital number, since it uses as much as digits as its factorization.
120231533413471 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 181440, while the sum is 40.
Adding to 120231533413471 its reverse (174314335132021), we get a palindrome (294545868545492).
The spelling of 120231533413471 in words is "one hundred twenty trillion, two hundred thirty-one billion, five hundred thirty-three million, four hundred thirteen thousand, four hundred seventy-one".
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