Base | Representation |
---|---|
bin | 100111000101001010110… |
… | …1001011010111011111001 |
3 | 201000111001112122000022111 |
4 | 1032022111221122323321 |
5 | 1201000224231032132 |
6 | 15231300532452321 |
7 | 1063025555600155 |
oct | 116122551327371 |
9 | 21014045560274 |
10 | 5371219783417 |
11 | 1790a155a41a4 |
12 | 728b8a77a0a1 |
13 | 2cc671c6a647 |
14 | 147d7b500065 |
15 | 94ab7638647 |
hex | 4e295a5aef9 |
5371219783417 has 2 divisors, whose sum is σ = 5371219783418. Its totient is φ = 5371219783416.
The previous prime is 5371219783399. The next prime is 5371219783429. The reversal of 5371219783417 is 7143879121735.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 5017134090816 + 354085692601 = 2239896^2 + 595051^2 .
It is a cyclic number.
It is not a de Polignac number, because 5371219783417 - 27 = 5371219783289 is a prime.
It is not a weakly prime, because it can be changed into another prime (5371219783817) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2685609891708 + 2685609891709.
It is an arithmetic number, because the mean of its divisors is an integer number (2685609891709).
Almost surely, 25371219783417 is an apocalyptic number.
It is an amenable number.
5371219783417 is a deficient number, since it is larger than the sum of its proper divisors (1).
5371219783417 is an equidigital number, since it uses as much as digits as its factorization.
5371219783417 is an evil number, because the sum of its binary digits is even.
The product of its digits is 8890560, while the sum is 58.
The spelling of 5371219783417 in words is "five trillion, three hundred seventy-one billion, two hundred nineteen million, seven hundred eighty-three thousand, four hundred seventeen".
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