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41550301510433 is a prime number
BaseRepresentation
bin10010111001010001011101…
…11101100011101100100001
312110010011120221201122012222
421130220232331203230201
520421230004141313213
6224211540042000425
711515624130335142
oct1134505675435441
9173104527648188
1041550301510433
111226a424a96791
1247b0887aa6115
131a2523b05a211
14a3908c51dac9
154c0c43167808
hex25ca2ef63b21

41550301510433 has 2 divisors, whose sum is σ = 41550301510434. Its totient is φ = 41550301510432.

The previous prime is 41550301510381. The next prime is 41550301510453. The reversal of 41550301510433 is 33401510305514.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 37512966797089 + 4037334713344 = 6124783^2 + 2009312^2 .

It is a cyclic number.

It is not a de Polignac number, because 41550301510433 - 244 = 23958115466017 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 41550301510393 and 41550301510402.

It is not a weakly prime, because it can be changed into another prime (41550301510453) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 20775150755216 + 20775150755217.

It is an arithmetic number, because the mean of its divisors is an integer number (20775150755217).

Almost surely, 241550301510433 is an apocalyptic number.

It is an amenable number.

41550301510433 is a deficient number, since it is larger than the sum of its proper divisors (1).

41550301510433 is an equidigital number, since it uses as much as digits as its factorization.

41550301510433 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 54000, while the sum is 35.

Adding to 41550301510433 its reverse (33401510305514), we get a palindrome (74951811815947).

The spelling of 41550301510433 in words is "forty-one trillion, five hundred fifty billion, three hundred one million, five hundred ten thousand, four hundred thirty-three".