Base | Representation |
---|---|
bin | 11101110010011110100100… |
… | …010001001111101100100001 |
3 | 122011212122020112102001200202 |
4 | 131302132210101033230201 |
5 | 114133000332203222144 |
6 | 1142350051531554545 |
7 | 36411213021604142 |
oct | 3562364421175441 |
9 | 564778215361622 |
10 | 131012143414049 |
11 | 38820a355288a7 |
12 | 1283b0458bba55 |
13 | 5814513619811 |
14 | 244d0432b5ac9 |
15 | 1022dd6e5d64e |
hex | 7727a444fb21 |
131012143414049 has 2 divisors, whose sum is σ = 131012143414050. Its totient is φ = 131012143414048.
The previous prime is 131012143414001. The next prime is 131012143414063. The reversal of 131012143414049 is 940414341210131.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 130279236204049 + 732907210000 = 11413993^2 + 856100^2 .
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-131012143414049 is a prime.
It is not a weakly prime, because it can be changed into another prime (131012143414069) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65506071707024 + 65506071707025.
It is an arithmetic number, because the mean of its divisors is an integer number (65506071707025).
Almost surely, 2131012143414049 is an apocalyptic number.
It is an amenable number.
131012143414049 is a deficient number, since it is larger than the sum of its proper divisors (1).
131012143414049 is an equidigital number, since it uses as much as digits as its factorization.
131012143414049 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 41472, while the sum is 38.
The spelling of 131012143414049 in words is "one hundred thirty-one trillion, twelve billion, one hundred forty-three million, four hundred fourteen thousand, forty-nine".
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