Base | Representation |
---|---|
bin | 11101110010011100101101… |
… | …000110100001011010101011 |
3 | 122011212110002012101010002122 |
4 | 131302130231012201122223 |
5 | 114132432233343124342 |
6 | 1142345121312021455 |
7 | 36411112334050061 |
oct | 3562345506413253 |
9 | 564773065333078 |
10 | 131010144114347 |
11 | 388200a9a1a3a3 |
12 | 1283a78823188b |
13 | 5814285359144 |
14 | 244ccd3960031 |
15 | 1022d1b6884d2 |
hex | 77272d1a16ab |
131010144114347 has 2 divisors, whose sum is σ = 131010144114348. Its totient is φ = 131010144114346.
The previous prime is 131010144114319. The next prime is 131010144114467. The reversal of 131010144114347 is 743411441010131.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 131010144114347 - 210 = 131010144113323 is a prime.
It is a super-2 number, since 2×1310101441143472 (a number of 29 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (131010144114247) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65505072057173 + 65505072057174.
It is an arithmetic number, because the mean of its divisors is an integer number (65505072057174).
Almost surely, 2131010144114347 is an apocalyptic number.
131010144114347 is a deficient number, since it is larger than the sum of its proper divisors (1).
131010144114347 is an equidigital number, since it uses as much as digits as its factorization.
131010144114347 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 35.
Adding to 131010144114347 its reverse (743411441010131), we get a palindrome (874421585124478).
The spelling of 131010144114347 in words is "one hundred thirty-one trillion, ten billion, one hundred forty-four million, one hundred fourteen thousand, three hundred forty-seven".
• e-mail: info -at- numbersaplenty.com • Privacy notice • done in 0.068 sec. • engine limits •