Base | Representation |
---|---|
bin | 1001100101001011… |
… | …01111000111111001 |
3 | 111021110211001221111 |
4 | 10302211233013321 |
5 | 41013243023231 |
6 | 2210223440321 |
7 | 241315621336 |
oct | 46245570771 |
9 | 14243731844 |
10 | 5143720441 |
11 | 21aa550332 |
12 | bb67560a1 |
13 | 63c86c482 |
14 | 36b1d398d |
15 | 2018954b1 |
hex | 13296f1f9 |
5143720441 has 2 divisors, whose sum is σ = 5143720442. Its totient is φ = 5143720440.
The previous prime is 5143720423. The next prime is 5143720487. The reversal of 5143720441 is 1440273415.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 5140316416 + 3404025 = 71696^2 + 1845^2 .
It is a cyclic number.
It is not a de Polignac number, because 5143720441 - 25 = 5143720409 is a prime.
It is a super-2 number, since 2×51437204412 = 52915719950322468962, which contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (5143723441) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (19) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2571860220 + 2571860221.
It is an arithmetic number, because the mean of its divisors is an integer number (2571860221).
Almost surely, 25143720441 is an apocalyptic number.
It is an amenable number.
5143720441 is a deficient number, since it is larger than the sum of its proper divisors (1).
5143720441 is an equidigital number, since it uses as much as digits as its factorization.
5143720441 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 13440, while the sum is 31.
The square root of 5143720441 is about 71719.7353662156. The cubic root of 5143720441 is about 1726.2053261585.
Adding to 5143720441 its reverse (1440273415), we get a palindrome (6583993856).
The spelling of 5143720441 in words is "five billion, one hundred forty-three million, seven hundred twenty thousand, four hundred forty-one".
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