Search a number
-
+
12241223144113 is a prime number
BaseRepresentation
bin1011001000100010000111…
…0000011101001010110001
31121100020201212222111002011
42302020201300131022301
53101030011111102423
642011312411122521
72402253630631105
oct262104160351261
947306655874064
1012241223144113
11399a523378347
121458520a43441
136aa45b2ca012
1430469a8a3905
15163650439b0d
hexb2221c1d2b1

12241223144113 has 2 divisors, whose sum is σ = 12241223144114. Its totient is φ = 12241223144112.

The previous prime is 12241223144077. The next prime is 12241223144119. The reversal of 12241223144113 is 31144132214221.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 6462257157409 + 5778965986704 = 2542097^2 + 2403948^2 .

It is a cyclic number.

It is not a de Polignac number, because 12241223144113 - 29 = 12241223143601 is a prime.

It is a super-4 number, since 4×122412231441134 (a number of 53 digits) contains 4444 as substring. Note that it is a super-d number also for d = 2.

It is not a weakly prime, because it can be changed into another prime (12241223144119) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6120611572056 + 6120611572057.

It is an arithmetic number, because the mean of its divisors is an integer number (6120611572057).

Almost surely, 212241223144113 is an apocalyptic number.

It is an amenable number.

12241223144113 is a deficient number, since it is larger than the sum of its proper divisors (1).

12241223144113 is an equidigital number, since it uses as much as digits as its factorization.

12241223144113 is an evil number, because the sum of its binary digits is even.

The product of its digits is 9216, while the sum is 31.

Adding to 12241223144113 its reverse (31144132214221), we get a palindrome (43385355358334).

The spelling of 12241223144113 in words is "twelve trillion, two hundred forty-one billion, two hundred twenty-three million, one hundred forty-four thousand, one hundred thirteen".