Base | Representation |
---|---|
bin | 10100110101111000010… |
… | …010011010001011011011 |
3 | 12001220212102101022210211 |
4 | 110311320102122023123 |
5 | 141431213000420212 |
6 | 3013544000551551 |
7 | 205322203561651 |
oct | 24657022321333 |
9 | 5056772338724 |
10 | 1432242201307 |
11 | 502457425355 |
12 | 1b16b30225b7 |
13 | a50a16880b8 |
14 | 4d46c758dd1 |
15 | 273c8a3e9a7 |
hex | 14d7849a2db |
1432242201307 has 2 divisors, whose sum is σ = 1432242201308. Its totient is φ = 1432242201306.
The previous prime is 1432242201293. The next prime is 1432242201317. The reversal of 1432242201307 is 7031022422341.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 1432242201307 - 27 = 1432242201179 is a prime.
It is a super-2 number, since 2×14322422013072 (a number of 25 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (1432242201317) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 716121100653 + 716121100654.
It is an arithmetic number, because the mean of its divisors is an integer number (716121100654).
Almost surely, 21432242201307 is an apocalyptic number.
1432242201307 is a deficient number, since it is larger than the sum of its proper divisors (1).
1432242201307 is an equidigital number, since it uses as much as digits as its factorization.
1432242201307 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 16128, while the sum is 31.
Adding to 1432242201307 its reverse (7031022422341), we get a palindrome (8463264623648).
The spelling of 1432242201307 in words is "one trillion, four hundred thirty-two billion, two hundred forty-two million, two hundred one thousand, three hundred seven".
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