Base | Representation |
---|---|
bin | 100110001100111000011… |
… | …1001110000011011010111 |
3 | 200120221001111221101201221 |
4 | 1030121300321300123113 |
5 | 1142010201110140341 |
6 | 15055550322031211 |
7 | 1051216265362603 |
oct | 114316071603327 |
9 | 20527044841657 |
10 | 5250344224471 |
11 | 1744727490745 |
12 | 7096757a8507 |
13 | 2c114a566578 |
14 | 142191ba0503 |
15 | 91890882ed1 |
hex | 4c670e706d7 |
5250344224471 has 2 divisors, whose sum is σ = 5250344224472. Its totient is φ = 5250344224470.
The previous prime is 5250344224441. The next prime is 5250344224493. The reversal of 5250344224471 is 1744224430525.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 5250344224471 - 25 = 5250344224439 is a prime.
It is a super-3 number, since 3×52503442244713 (a number of 39 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (5250344224441) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2625172112235 + 2625172112236.
It is an arithmetic number, because the mean of its divisors is an integer number (2625172112236).
Almost surely, 25250344224471 is an apocalyptic number.
5250344224471 is a deficient number, since it is larger than the sum of its proper divisors (1).
5250344224471 is an equidigital number, since it uses as much as digits as its factorization.
5250344224471 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 1075200, while the sum is 43.
Adding to 5250344224471 its reverse (1744224430525), we get a palindrome (6994568654996).
The spelling of 5250344224471 in words is "five trillion, two hundred fifty billion, three hundred forty-four million, two hundred twenty-four thousand, four hundred seventy-one".
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