Base | Representation |
---|---|
bin | 11111001001011010101… |
… | …001001010111010011011 |
3 | 21120121210011102020201011 |
4 | 133021122221022322123 |
5 | 240032032100033301 |
6 | 4315143130034351 |
7 | 310432315453216 |
oct | 37113251127233 |
9 | 7517704366634 |
10 | 2140414455451 |
11 | 755821183498 |
12 | 2a69b05789b7 |
13 | 126abcc5947c |
14 | 7584d16677d |
15 | 3aa252baa51 |
hex | 1f25aa4ae9b |
2140414455451 has 2 divisors, whose sum is σ = 2140414455452. Its totient is φ = 2140414455450.
The previous prime is 2140414455341. The next prime is 2140414455497. The reversal of 2140414455451 is 1545544140412.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 2140414455451 - 215 = 2140414422683 is a prime.
It is a super-3 number, since 3×21404144554513 (a number of 38 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.
It is not a weakly prime, because it can be changed into another prime (2140444455451) 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, 1070207227725 + 1070207227726.
It is an arithmetic number, because the mean of its divisors is an integer number (1070207227726).
Almost surely, 22140414455451 is an apocalyptic number.
2140414455451 is a deficient number, since it is larger than the sum of its proper divisors (1).
2140414455451 is an equidigital number, since it uses as much as digits as its factorization.
2140414455451 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 256000, while the sum is 40.
Adding to 2140414455451 its reverse (1545544140412), we get a palindrome (3685958595863).
The spelling of 2140414455451 in words is "two trillion, one hundred forty billion, four hundred fourteen million, four hundred fifty-five thousand, four hundred fifty-one".
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