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201315253441 is a prime number
BaseRepresentation
bin1011101101111101010…
…0101111110011000001
3201020122000011111000211
42323133110233303001
511244243201102231
6232252152314121
720354531264101
oct2733724576301
9636560144024
10201315253441
1178417237615
12330240a2941
1315ca3967c53
149a5aa53201
155383b78eb1
hex2edf52fcc1

201315253441 has 2 divisors, whose sum is σ = 201315253442. Its totient is φ = 201315253440.

The previous prime is 201315253357. The next prime is 201315253463. The reversal of 201315253441 is 144352513102.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 145939280400 + 55375973041 = 382020^2 + 235321^2 .

It is a cyclic number.

It is not a de Polignac number, because 201315253441 - 223 = 201306864833 is a prime.

It is not a weakly prime, because it can be changed into another prime (201315253471) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (23) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 100657626720 + 100657626721.

It is an arithmetic number, because the mean of its divisors is an integer number (100657626721).

Almost surely, 2201315253441 is an apocalyptic number.

It is an amenable number.

201315253441 is a deficient number, since it is larger than the sum of its proper divisors (1).

201315253441 is an equidigital number, since it uses as much as digits as its factorization.

201315253441 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 14400, while the sum is 31.

Adding to 201315253441 its reverse (144352513102), we get a palindrome (345667766543).

The spelling of 201315253441 in words is "two hundred one billion, three hundred fifteen million, two hundred fifty-three thousand, four hundred forty-one".