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144214022131 is a prime number
BaseRepresentation
bin1000011001001111010…
…0110101011111110011
3111210020112010101100221
42012103310311133303
54330322242202011
6150130114553511
713263520651054
oct2062364653763
9453215111327
10144214022131
1156185068363
1223b48b98897
13107a393a20a
146da117472b
153b40b70971
hex2193d357f3

144214022131 has 2 divisors, whose sum is σ = 144214022132. Its totient is φ = 144214022130.

The previous prime is 144214022123. The next prime is 144214022177. The reversal of 144214022131 is 131220412441.

It is a weak prime.

It is a cyclic number.

It is not a de Polignac number, because 144214022131 - 23 = 144214022123 is a prime.

It is a super-3 number, since 3×1442140221313 (a number of 34 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.

It is a junction number, because it is equal to n+sod(n) for n = 144214022096 and 144214022105.

It is not a weakly prime, because it can be changed into another prime (144214023131) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 72107011065 + 72107011066.

It is an arithmetic number, because the mean of its divisors is an integer number (72107011066).

Almost surely, 2144214022131 is an apocalyptic number.

144214022131 is a deficient number, since it is larger than the sum of its proper divisors (1).

144214022131 is an equidigital number, since it uses as much as digits as its factorization.

144214022131 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 1536, while the sum is 25.

Adding to 144214022131 its reverse (131220412441), we get a palindrome (275434434572).

The spelling of 144214022131 in words is "one hundred forty-four billion, two hundred fourteen million, twenty-two thousand, one hundred thirty-one".