Base | Representation |
---|---|
bin | 100101010000010000111… |
… | …1110111001100000111111 |
3 | 200010111001112100020100002 |
4 | 1022200201332321200333 |
5 | 1132342102303402341 |
6 | 14520101341224515 |
7 | 1035630420466223 |
oct | 112404176714077 |
9 | 20114045306302 |
10 | 5120171153471 |
11 | 16a44a8151830 |
12 | 6a83a6b6213b |
13 | 2b1aa387597a |
14 | 139b63722783 |
15 | 8d2c273429b |
hex | 4a821fb983f |
5120171153471 has 16 divisors (see below), whose sum is σ = 5589971598240. Its totient is φ = 4651093332480.
The previous prime is 5120171153467. The next prime is 5120171153473. The reversal of 5120171153471 is 1743511710215.
It is a cyclic number.
It is not a de Polignac number, because 5120171153471 - 22 = 5120171153467 is a prime.
It is a super-3 number, since 3×51201711534713 (a number of 39 digits) contains 333 as substring.
It is a Duffinian number.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (5120171153473) by changing a digit.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 145735835 + ... + 145770963.
It is an arithmetic number, because the mean of its divisors is an integer number (349373224890).
Almost surely, 25120171153471 is an apocalyptic number.
5120171153471 is a deficient number, since it is larger than the sum of its proper divisors (469800444769).
5120171153471 is a wasteful number, since it uses less digits than its factorization.
5120171153471 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 45034.
The product of its (nonzero) digits is 29400, while the sum is 38.
Adding to 5120171153471 its reverse (1743511710215), we get a palindrome (6863682863686).
The spelling of 5120171153471 in words is "five trillion, one hundred twenty billion, one hundred seventy-one million, one hundred fifty-three thousand, four hundred seventy-one".
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