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5268640785433 is a prime number
BaseRepresentation
bin100110010101011001101…
…1101100110010000011001
3200122200021122010222220221
41030222303131212100121
51142310134020113213
615112214040520041
71052434554522124
oct114526335462031
920580248128827
105268640785433
111751466408803
12711121180021
132c2aa505c4bc
14143009b473bb
15920b1cbb38d
hex4cab3766419

5268640785433 has 2 divisors, whose sum is σ = 5268640785434. Its totient is φ = 5268640785432.

The previous prime is 5268640785361. The next prime is 5268640785461. The reversal of 5268640785433 is 3345870468625.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 4051255149529 + 1217385635904 = 2012773^2 + 1103352^2 .

It is a cyclic number.

It is not a de Polignac number, because 5268640785433 - 225 = 5268607231001 is a prime.

It is a super-3 number, since 3×52686407854333 (a number of 39 digits) contains 333 as substring.

It is not a weakly prime, because it can be changed into another prime (5268640783433) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2634320392716 + 2634320392717.

It is an arithmetic number, because the mean of its divisors is an integer number (2634320392717).

Almost surely, 25268640785433 is an apocalyptic number.

It is an amenable number.

5268640785433 is a deficient number, since it is larger than the sum of its proper divisors (1).

5268640785433 is an equidigital number, since it uses as much as digits as its factorization.

5268640785433 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 116121600, while the sum is 61.

The spelling of 5268640785433 in words is "five trillion, two hundred sixty-eight billion, six hundred forty million, seven hundred eighty-five thousand, four hundred thirty-three".