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43275212131 is a prime number
BaseRepresentation
bin101000010011011001…
…110101010101100011
311010200220221112220111
4220103121311111203
51202111423242011
631514053552151
73061254143134
oct502331652543
9133626845814
1043275212131
1117397802428
128478945657
1341087960c8
142147573c8b
1511d42e0121
hexa13675563

43275212131 has 2 divisors, whose sum is σ = 43275212132. Its totient is φ = 43275212130.

The previous prime is 43275212119. The next prime is 43275212143. The reversal of 43275212131 is 13121257234.

43275212131 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a balanced prime because it is at equal distance from previous prime (43275212119) and next prime (43275212143).

It is a cyclic number.

It is not a de Polignac number, because 43275212131 - 29 = 43275211619 is a prime.

It is a super-2 number, since 2×432752121312 (a number of 22 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 43275212093 and 43275212102.

It is not a weakly prime, because it can be changed into another prime (43275212111) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 21637606065 + 21637606066.

It is an arithmetic number, because the mean of its divisors is an integer number (21637606066).

Almost surely, 243275212131 is an apocalyptic number.

43275212131 is a deficient number, since it is larger than the sum of its proper divisors (1).

43275212131 is an equidigital number, since it uses as much as digits as its factorization.

43275212131 is an evil number, because the sum of its binary digits is even.

The product of its digits is 10080, while the sum is 31.

Adding to 43275212131 its reverse (13121257234), we get a palindrome (56396469365).

The spelling of 43275212131 in words is "forty-three billion, two hundred seventy-five million, two hundred twelve thousand, one hundred thirty-one".