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43141421131 is a prime number
BaseRepresentation
bin101000001011011011…
…011101100001001011
311010100121012021011021
4220023123131201023
51201323200434011
631452514213311
73055041012664
oct501333354113
9133317167137
1043141421131
1117329231256
12843bb84237
1340b6b50c33
1421338a836b
1511c76b3471
hexa0b6dd84b

43141421131 has 2 divisors, whose sum is σ = 43141421132. Its totient is φ = 43141421130.

The previous prime is 43141421129. The next prime is 43141421153. The reversal of 43141421131 is 13112414134.

43141421131 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a weak prime.

It is a cyclic number.

It is not a de Polignac number, because 43141421131 - 21 = 43141421129 is a prime.

It is a super-2 number, since 2×431414211312 (a number of 22 digits) contains 22 as substring.

Together with 43141421129, it forms a pair of twin primes.

It is a junction number, because it is equal to n+sod(n) for n = 43141421096 and 43141421105.

It is not a weakly prime, because it can be changed into another prime (43141121131) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 21570710565 + 21570710566.

It is an arithmetic number, because the mean of its divisors is an integer number (21570710566).

Almost surely, 243141421131 is an apocalyptic number.

43141421131 is a deficient number, since it is larger than the sum of its proper divisors (1).

43141421131 is an equidigital number, since it uses as much as digits as its factorization.

43141421131 is an evil number, because the sum of its binary digits is even.

The product of its digits is 1152, while the sum is 25.

Adding to 43141421131 its reverse (13112414134), we get a palindrome (56253835265).

The spelling of 43141421131 in words is "forty-three billion, one hundred forty-one million, four hundred twenty-one thousand, one hundred thirty-one".