Base | Representation |
---|---|
bin | 11101111010100001100000… |
… | …010101000101101011011011 |
3 | 122020211111101211212220212012 |
4 | 131322201200111011223123 |
5 | 114221030212403231331 |
6 | 1143452052424140135 |
7 | 36466155460204223 |
oct | 3572414025055333 |
9 | 566744354786765 |
10 | 131565054352091 |
11 | 38a14476319168 |
12 | 1290a23260764b |
13 | 58546ba607849 |
14 | 246bad568b083 |
15 | 1032497c8682b |
hex | 77a860545adb |
131565054352091 has 2 divisors, whose sum is σ = 131565054352092. Its totient is φ = 131565054352090.
The previous prime is 131565054352049. The next prime is 131565054352133. The reversal of 131565054352091 is 190253450565131.
It is a balanced prime because it is at equal distance from previous prime (131565054352049) and next prime (131565054352133).
It is a cyclic number.
It is not a de Polignac number, because 131565054352091 - 218 = 131565054089947 is a prime.
It is not a weakly prime, because it can be changed into another prime (131565054352001) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65782527176045 + 65782527176046.
It is an arithmetic number, because the mean of its divisors is an integer number (65782527176046).
Almost surely, 2131565054352091 is an apocalyptic number.
131565054352091 is a deficient number, since it is larger than the sum of its proper divisors (1).
131565054352091 is an equidigital number, since it uses as much as digits as its factorization.
131565054352091 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 2430000, while the sum is 50.
The spelling of 131565054352091 in words is "one hundred thirty-one trillion, five hundred sixty-five billion, fifty-four million, three hundred fifty-two thousand, ninety-one".
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