Base | Representation |
---|---|
bin | 10110011101111001… |
… | …111010001110110001 |
3 | 2022021020122012202011 |
4 | 112131321322032301 |
5 | 343401223240423 |
6 | 15025445522521 |
7 | 1512546646315 |
oct | 263571721661 |
9 | 68236565664 |
10 | 24124040113 |
11 | a25a434438 |
12 | 4813113441 |
13 | 2375c09164 |
14 | 124bcc7b45 |
15 | 962d43b0d |
hex | 59de7a3b1 |
24124040113 has 2 divisors, whose sum is σ = 24124040114. Its totient is φ = 24124040112.
The previous prime is 24124040093. The next prime is 24124040117. The reversal of 24124040113 is 31104042142.
It is a happy number.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 15698085264 + 8425954849 = 125292^2 + 91793^2 .
It is a cyclic number.
It is not a de Polignac number, because 24124040113 - 25 = 24124040081 is a prime.
It is a super-2 number, since 2×241240401132 (a number of 22 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (24124040117) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 12062020056 + 12062020057.
It is an arithmetic number, because the mean of its divisors is an integer number (12062020057).
Almost surely, 224124040113 is an apocalyptic number.
It is an amenable number.
24124040113 is a deficient number, since it is larger than the sum of its proper divisors (1).
24124040113 is an equidigital number, since it uses as much as digits as its factorization.
24124040113 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 768, while the sum is 22.
Adding to 24124040113 its reverse (31104042142), we get a palindrome (55228082255).
The spelling of 24124040113 in words is "twenty-four billion, one hundred twenty-four million, forty thousand, one hundred thirteen".
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