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531311205011233 is a prime number
BaseRepresentation
bin111100011001110011000011…
…1101000111100111100100001
32120200012212002101211101220211
41320303212013220330330201
51024120000321440324413
65122000555322534121
7216624652650266653
oct17063460750747441
92520185071741824
10531311205011233
111443238a8a79879
124b70b807667341
1319a6057c444c1a
14952b85718a6d3
15416595518583d
hex1e33987a3cf21

531311205011233 has 2 divisors, whose sum is σ = 531311205011234. Its totient is φ = 531311205011232.

The previous prime is 531311205011227. The next prime is 531311205011297. The reversal of 531311205011233 is 332110502113135.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 493747642707489 + 37563562303744 = 22220433^2 + 6128912^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-531311205011233 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 531311205011195 and 531311205011204.

It is not a weakly prime, because it can be changed into another prime (531311205011633) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 265655602505616 + 265655602505617.

It is an arithmetic number, because the mean of its divisors is an integer number (265655602505617).

Almost surely, 2531311205011233 is an apocalyptic number.

It is an amenable number.

531311205011233 is a deficient number, since it is larger than the sum of its proper divisors (1).

531311205011233 is an equidigital number, since it uses as much as digits as its factorization.

531311205011233 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 8100, while the sum is 31.

Adding to 531311205011233 its reverse (332110502113135), we get a palindrome (863421707124368).

The spelling of 531311205011233 in words is "five hundred thirty-one trillion, three hundred eleven billion, two hundred five million, eleven thousand, two hundred thirty-three".