Base | Representation |
---|---|
bin | 10111010000001100101101… |
… | …01001100001000101001011 |
3 | 20201001101002021002022120221 |
4 | 23220012112221201011023 |
5 | 23200240130424442213 |
6 | 300430404241330511 |
7 | 13525213664055511 |
oct | 1350062651410513 |
9 | 221041067068527 |
10 | 51134113124683 |
11 | 15324945882606 |
12 | 589a173036a37 |
13 | 226bc06a72414 |
14 | c8ac934ba4b1 |
15 | 5da1b03a838d |
hex | 2e8196a6114b |
51134113124683 has 2 divisors, whose sum is σ = 51134113124684. Its totient is φ = 51134113124682.
The previous prime is 51134113124587. The next prime is 51134113124711. The reversal of 51134113124683 is 38642131143115.
It is a happy number.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-51134113124683 is a prime.
It is a super-2 number, since 2×511341131246832 (a number of 28 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (51134113184683) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 25567056562341 + 25567056562342.
It is an arithmetic number, because the mean of its divisors is an integer number (25567056562342).
Almost surely, 251134113124683 is an apocalyptic number.
51134113124683 is a deficient number, since it is larger than the sum of its proper divisors (1).
51134113124683 is an equidigital number, since it uses as much as digits as its factorization.
51134113124683 is an evil number, because the sum of its binary digits is even.
The product of its digits is 207360, while the sum is 43.
Adding to 51134113124683 its reverse (38642131143115), we get a palindrome (89776244267798).
The spelling of 51134113124683 in words is "fifty-one trillion, one hundred thirty-four billion, one hundred thirteen million, one hundred twenty-four thousand, six hundred eighty-three".
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