Base | Representation |
---|---|
bin | 1100111011001101001… |
… | …10100000111111100011 |
3 | 1120110022022011010001021 |
4 | 12131212212200333203 |
5 | 24234010332113011 |
6 | 540003524350311 |
7 | 44041200151663 |
oct | 6354646407743 |
9 | 1513268133037 |
10 | 444103004131 |
11 | 161385628146 |
12 | 720a1245397 |
13 | 32b567682a7 |
14 | 176cd6581a3 |
15 | b843700471 |
hex | 67669a0fe3 |
444103004131 has 2 divisors, whose sum is σ = 444103004132. Its totient is φ = 444103004130.
The previous prime is 444103004117. The next prime is 444103004143. The reversal of 444103004131 is 131400301444.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 444103004131 - 217 = 444102873059 is a prime.
It is a super-2 number, since 2×4441030041312 (a number of 24 digits) contains 22 as substring.
It is a junction number, because it is equal to n+sod(n) for n = 444103004096 and 444103004105.
It is not a weakly prime, because it can be changed into another prime (444103004161) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 222051502065 + 222051502066.
It is an arithmetic number, because the mean of its divisors is an integer number (222051502066).
Almost surely, 2444103004131 is an apocalyptic number.
444103004131 is a deficient number, since it is larger than the sum of its proper divisors (1).
444103004131 is an equidigital number, since it uses as much as digits as its factorization.
444103004131 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 2304, while the sum is 25.
Adding to 444103004131 its reverse (131400301444), we get a palindrome (575503305575).
The spelling of 444103004131 in words is "four hundred forty-four billion, one hundred three million, four thousand, one hundred thirty-one".
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