Base | Representation |
---|---|
bin | 1011000001111011001101… |
… | …0001010111100010101011 |
3 | 1120221101201021211002020112 |
4 | 2300132303101113202223 |
5 | 3042200012242214042 |
6 | 41443215535444535 |
7 | 2361124502050163 |
oct | 260366321274253 |
9 | 46841637732215 |
10 | 12127700351147 |
11 | 3956368752797 |
12 | 143a51a38674b |
13 | 69c839013044 |
14 | 2dcdab8ab5a3 |
15 | 160708e31982 |
hex | b07b34578ab |
12127700351147 has 2 divisors, whose sum is σ = 12127700351148. Its totient is φ = 12127700351146.
The previous prime is 12127700351119. The next prime is 12127700351153. The reversal of 12127700351147 is 74115300772121.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-12127700351147 is a prime.
It is a super-2 number, since 2×121277003511472 (a number of 27 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (12127700351167) 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, 6063850175573 + 6063850175574.
It is an arithmetic number, because the mean of its divisors is an integer number (6063850175574).
Almost surely, 212127700351147 is an apocalyptic number.
12127700351147 is a deficient number, since it is larger than the sum of its proper divisors (1).
12127700351147 is an equidigital number, since it uses as much as digits as its factorization.
12127700351147 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 82320, while the sum is 41.
The spelling of 12127700351147 in words is "twelve trillion, one hundred twenty-seven billion, seven hundred million, three hundred fifty-one thousand, one hundred forty-seven".
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