Base | Representation |
---|---|
bin | 10010111010110101001… |
… | …101000011011000111111 |
3 | 11121021211121120022202221 |
4 | 102322311031003120333 |
5 | 132300121234034442 |
6 | 2433133400230211 |
7 | 162634122353551 |
oct | 22726515033077 |
9 | 4537747508687 |
10 | 1300120221247 |
11 | 461417a751a1 |
12 | 18bb7b76b367 |
13 | 957a6c9ac83 |
14 | 46cd75214d1 |
15 | 23c44771467 |
hex | 12eb534363f |
1300120221247 has 2 divisors, whose sum is σ = 1300120221248. Its totient is φ = 1300120221246.
The previous prime is 1300120221221. The next prime is 1300120221251. The reversal of 1300120221247 is 7421220210031.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 1300120221247 - 231 = 1297972737599 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (1300120221217) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (23) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 650060110623 + 650060110624.
It is an arithmetic number, because the mean of its divisors is an integer number (650060110624).
Almost surely, 21300120221247 is an apocalyptic number.
1300120221247 is a deficient number, since it is larger than the sum of its proper divisors (1).
1300120221247 is an equidigital number, since it uses as much as digits as its factorization.
1300120221247 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 1344, while the sum is 25.
Adding to 1300120221247 its reverse (7421220210031), we get a palindrome (8721340431278).
The spelling of 1300120221247 in words is "one trillion, three hundred billion, one hundred twenty million, two hundred twenty-one thousand, two hundred forty-seven".
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