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5431255035433 is a prime number
BaseRepresentation
bin100111100001001000000…
…0001101110111000101001
3201020020000111112210201221
41033002100001232320221
51202441202242113213
615315030100342041
71100252364036343
oct117022001567051
921206014483657
105431255035433
111804422969657
127387443a4921
1330521babb1c6
1414ac3492c293
159642d00138d
hex4f09006ee29

5431255035433 has 2 divisors, whose sum is σ = 5431255035434. Its totient is φ = 5431255035432.

The previous prime is 5431255035379. The next prime is 5431255035469. The reversal of 5431255035433 is 3345305521345.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 4309663896729 + 1121591138704 = 2075973^2 + 1059052^2 .

It is a cyclic number.

It is not a de Polignac number, because 5431255035433 - 225 = 5431221481001 is a prime.

It is a self number, because there is not a number n which added to its sum of digits gives 5431255035433.

It is not a weakly prime, because it can be changed into another prime (5431255033433) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2715627517716 + 2715627517717.

It is an arithmetic number, because the mean of its divisors is an integer number (2715627517717).

Almost surely, 25431255035433 is an apocalyptic number.

It is an amenable number.

5431255035433 is a deficient number, since it is larger than the sum of its proper divisors (1).

5431255035433 is an equidigital number, since it uses as much as digits as its factorization.

5431255035433 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 1620000, while the sum is 43.

The spelling of 5431255035433 in words is "five trillion, four hundred thirty-one billion, two hundred fifty-five million, thirty-five thousand, four hundred thirty-three".