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40325144332021 is a prime number
BaseRepresentation
bin10010010101100111011011…
…11001110100011011110101
312021210001020122100110011221
421022303231321310123311
520241141343422111041
6221433040543430341
711331252445212463
oct1112635571643365
9167701218313157
1040325144332021
111193788577915a
1246333492b53b1
13196684c7c5933
149d5a66c15833
1549de39aad2d1
hex24acede746f5

40325144332021 has 2 divisors, whose sum is σ = 40325144332022. Its totient is φ = 40325144332020.

The previous prime is 40325144332001. The next prime is 40325144332027. The reversal of 40325144332021 is 12023344152304.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 31307225421796 + 9017918910225 = 5595286^2 + 3002985^2 .

It is a cyclic number.

It is not a de Polignac number, because 40325144332021 - 27 = 40325144331893 is a prime.

It is a super-2 number, since 2×403251443320212 (a number of 28 digits) contains 22 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (40325144332027) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 20162572166010 + 20162572166011.

It is an arithmetic number, because the mean of its divisors is an integer number (20162572166011).

Almost surely, 240325144332021 is an apocalyptic number.

It is an amenable number.

40325144332021 is a deficient number, since it is larger than the sum of its proper divisors (1).

40325144332021 is an equidigital number, since it uses as much as digits as its factorization.

40325144332021 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 69120, while the sum is 34.

Adding to 40325144332021 its reverse (12023344152304), we get a palindrome (52348488484325).

The spelling of 40325144332021 in words is "forty trillion, three hundred twenty-five billion, one hundred forty-four million, three hundred thirty-two thousand, twenty-one".