Base | Representation |
---|---|
bin | 10100100010100000100… |
… | …000000100001110000111 |
3 | 11222221011211012022201222 |
4 | 110202200200010032013 |
5 | 141111113130033133 |
6 | 3000224005110555 |
7 | 203654564222303 |
oct | 24424040041607 |
9 | 4887154168658 |
10 | 1411442033543 |
11 | 4a4653260463 |
12 | 1a96690b845b |
13 | a313829c119 |
14 | 4c4580a2d03 |
15 | 26aac916e98 |
hex | 148a0804387 |
1411442033543 has 2 divisors, whose sum is σ = 1411442033544. Its totient is φ = 1411442033542.
The previous prime is 1411442033537. The next prime is 1411442033567. The reversal of 1411442033543 is 3453302441141.
It is a weak prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-1411442033543 is a prime.
It is a super-2 number, since 2×14114420335432 (a number of 25 digits) contains 22 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (1411442033513) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (13) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 705721016771 + 705721016772.
It is an arithmetic number, because the mean of its divisors is an integer number (705721016772).
Almost surely, 21411442033543 is an apocalyptic number.
1411442033543 is a deficient number, since it is larger than the sum of its proper divisors (1).
1411442033543 is an equidigital number, since it uses as much as digits as its factorization.
1411442033543 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 69120, while the sum is 35.
Adding to 1411442033543 its reverse (3453302441141), we get a palindrome (4864744474684).
The spelling of 1411442033543 in words is "one trillion, four hundred eleven billion, four hundred forty-two million, thirty-three thousand, five hundred forty-three".
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