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12471272594213 is a prime number
BaseRepresentation
bin1011010101111011000111…
…0001011101101100100101
31122011020112102121101201212
42311132301301131230211
53113312131241003323
642305120141201205
72425006521661256
oct265366161355445
948136472541655
1012471272594213
113a79045288124
121495023b60205
136c60610062a4
143118817c2b2d
15169616a04b78
hexb57b1c5db25

12471272594213 has 2 divisors, whose sum is σ = 12471272594214. Its totient is φ = 12471272594212.

The previous prime is 12471272594161. The next prime is 12471272594219. The reversal of 12471272594213 is 31249527217421.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 10595558826724 + 1875713767489 = 3255082^2 + 1369567^2 .

It is a cyclic number.

It is not a de Polignac number, because 12471272594213 - 218 = 12471272332069 is a prime.

It is a super-2 number, since 2×124712725942132 (a number of 27 digits) contains 22 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (12471272594219) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6235636297106 + 6235636297107.

It is an arithmetic number, because the mean of its divisors is an integer number (6235636297107).

Almost surely, 212471272594213 is an apocalyptic number.

It is an amenable number.

12471272594213 is a deficient number, since it is larger than the sum of its proper divisors (1).

12471272594213 is an equidigital number, since it uses as much as digits as its factorization.

12471272594213 is an odious number, because the sum of its binary digits is odd.

The product of its digits is 1693440, while the sum is 50.

The spelling of 12471272594213 in words is "twelve trillion, four hundred seventy-one billion, two hundred seventy-two million, five hundred ninety-four thousand, two hundred thirteen".