Search a number
-
+
5320443112973 is a prime number
BaseRepresentation
bin100110101101100001100…
…0111110010111000001101
3200211121222211100220010122
41031123003013302320031
51144132231414103343
615152102224314325
71056250353165122
oct115330307627015
920747884326118
105320443112973
11177142948514a
1271b1797729a5
132c793c325811
14145721998749
15935e49c9068
hex4d6c31f2e0d

5320443112973 has 2 divisors, whose sum is σ = 5320443112974. Its totient is φ = 5320443112972.

The previous prime is 5320443112969. The next prime is 5320443112981. The reversal of 5320443112973 is 3792113440235.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 4485233558569 + 835209554404 = 2117837^2 + 913898^2 .

It is a cyclic number.

It is not a de Polignac number, because 5320443112973 - 22 = 5320443112969 is a prime.

It is a super-3 number, since 3×53204431129733 (a number of 39 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (5320443112373) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2660221556486 + 2660221556487.

It is an arithmetic number, because the mean of its divisors is an integer number (2660221556487).

Almost surely, 25320443112973 is an apocalyptic number.

It is an amenable number.

5320443112973 is a deficient number, since it is larger than the sum of its proper divisors (1).

5320443112973 is an equidigital number, since it uses as much as digits as its factorization.

5320443112973 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 544320, while the sum is 44.

The spelling of 5320443112973 in words is "five trillion, three hundred twenty billion, four hundred forty-three million, one hundred twelve thousand, nine hundred seventy-three".