Base | Representation |
---|---|
bin | 111101001010010111… |
… | …0100011010100000001 |
3 | 110120000121101020010001 |
4 | 1322110232203110001 |
5 | 4122443112000423 |
6 | 140201055040001 |
7 | 12326555463616 |
oct | 1722456432401 |
9 | 416017336101 |
10 | 131344250113 |
11 | 5078045013a |
12 | 21556b10001 |
13 | c5025339bb |
14 | 64ddc2850d |
15 | 363ad94cad |
hex | 1e94ba3501 |
131344250113 has 2 divisors, whose sum is σ = 131344250114. Its totient is φ = 131344250112.
The previous prime is 131344250077. The next prime is 131344250123. The reversal of 131344250113 is 311052443131.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 131009250304 + 334999809 = 361952^2 + 18303^2 .
It is an emirp because it is prime and its reverse (311052443131) is a distict prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-131344250113 is a prime.
It is not a weakly prime, because it can be changed into another prime (131344250123) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (17) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65672125056 + 65672125057.
It is an arithmetic number, because the mean of its divisors is an integer number (65672125057).
Almost surely, 2131344250113 is an apocalyptic number.
It is an amenable number.
131344250113 is a deficient number, since it is larger than the sum of its proper divisors (1).
131344250113 is an equidigital number, since it uses as much as digits as its factorization.
131344250113 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 4320, while the sum is 28.
Adding to 131344250113 its reverse (311052443131), we get a palindrome (442396693244).
The spelling of 131344250113 in words is "one hundred thirty-one billion, three hundred forty-four million, two hundred fifty thousand, one hundred thirteen".
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