Base | Representation |
---|---|
bin | 11110000010010111110100… |
… | …101100001000000111010011 |
3 | 122022202001120202211202012012 |
4 | 132002113310230020013103 |
5 | 114303344320032100342 |
6 | 1144543532524253135 |
7 | 36553141352322653 |
oct | 3602276454100723 |
9 | 568661522752165 |
10 | 132104414331347 |
11 | 39102192a31637 |
12 | 129968790a11ab |
13 | 5893515039173 |
14 | 2489c60078363 |
15 | 10415140cab82 |
hex | 7825f4b081d3 |
132104414331347 has 2 divisors, whose sum is σ = 132104414331348. Its totient is φ = 132104414331346.
The previous prime is 132104414331311. The next prime is 132104414331367. The reversal of 132104414331347 is 743133414401231.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 132104414331347 - 222 = 132104410137043 is a prime.
It is a super-2 number, since 2×1321044143313472 (a number of 29 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (132104414331367) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 66052207165673 + 66052207165674.
It is an arithmetic number, because the mean of its divisors is an integer number (66052207165674).
Almost surely, 2132104414331347 is an apocalyptic number.
132104414331347 is a deficient number, since it is larger than the sum of its proper divisors (1).
132104414331347 is an equidigital number, since it uses as much as digits as its factorization.
132104414331347 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 290304, while the sum is 41.
Adding to 132104414331347 its reverse (743133414401231), we get a palindrome (875237828732578).
The spelling of 132104414331347 in words is "one hundred thirty-two trillion, one hundred four billion, four hundred fourteen million, three hundred thirty-one thousand, three hundred forty-seven".
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