Base | Representation |
---|---|
bin | 11101110010011110101001… |
… | …000101011101010010000011 |
3 | 122011212122110011102220222021 |
4 | 131302132221011131102003 |
5 | 114133001013344210201 |
6 | 1142350103535422311 |
7 | 36411215022424042 |
oct | 3562365105352203 |
9 | 564778404386867 |
10 | 131012224210051 |
11 | 38820a770a3058 |
12 | 1283b068984997 |
13 | 58145272982c2 |
14 | 244d04dd08559 |
15 | 1022dde0bcea1 |
hex | 7727a915d483 |
131012224210051 has 2 divisors, whose sum is σ = 131012224210052. Its totient is φ = 131012224210050.
The previous prime is 131012224210027. The next prime is 131012224210069. The reversal of 131012224210051 is 150012422210131.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 131012224210051 - 227 = 131012089992323 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 131012224209998 and 131012224210025.
It is not a weakly prime, because it can be changed into another prime (131012224210081) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65506112105025 + 65506112105026.
It is an arithmetic number, because the mean of its divisors is an integer number (65506112105026).
Almost surely, 2131012224210051 is an apocalyptic number.
131012224210051 is a deficient number, since it is larger than the sum of its proper divisors (1).
131012224210051 is an equidigital number, since it uses as much as digits as its factorization.
131012224210051 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 960, while the sum is 25.
Adding to 131012224210051 its reverse (150012422210131), we get a palindrome (281024646420182).
The spelling of 131012224210051 in words is "one hundred thirty-one trillion, twelve billion, two hundred twenty-four million, two hundred ten thousand, fifty-one".
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