Base | Representation |
---|---|
bin | 11101110010010111000110… |
… | …110000100000100010101011 |
3 | 122011211212120012201011012111 |
4 | 131302113012300200202223 |
5 | 114132332430304202032 |
6 | 1142342244545423151 |
7 | 36410510342553166 |
oct | 3562270660404253 |
9 | 564755505634174 |
10 | 131004132100267 |
11 | 388185a4333218 |
12 | 1283958a93aab7 |
13 | 5813837944b6a |
14 | 244c8c3304add |
15 | 1022ab897ac47 |
hex | 7725c6c208ab |
131004132100267 has 2 divisors, whose sum is σ = 131004132100268. Its totient is φ = 131004132100266.
The previous prime is 131004132100253. The next prime is 131004132100277. The reversal of 131004132100267 is 762001231400131.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 131004132100267 - 231 = 131001984616619 is a prime.
It is not a weakly prime, because it can be changed into another prime (131004132100207) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65502066050133 + 65502066050134.
It is an arithmetic number, because the mean of its divisors is an integer number (65502066050134).
Almost surely, 2131004132100267 is an apocalyptic number.
131004132100267 is a deficient number, since it is larger than the sum of its proper divisors (1).
131004132100267 is an equidigital number, since it uses as much as digits as its factorization.
131004132100267 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 6048, while the sum is 31.
Adding to 131004132100267 its reverse (762001231400131), we get a palindrome (893005363500398).
The spelling of 131004132100267 in words is "one hundred thirty-one trillion, four billion, one hundred thirty-two million, one hundred thousand, two hundred sixty-seven".
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