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55055000500433 is a prime number
BaseRepresentation
bin11001000010010011111011…
…01111110111100011010001
321012221012120122102022012122
430201021331233313203101
524204010120112003213
6313031530145304025
714411410143626423
oct1441117557674321
9235835518368178
1055055000500433
11165a677055768a
126212040335015
13249488352c54b
14d84967872813
15657190c2ad08
hex32127dbf78d1

55055000500433 has 2 divisors, whose sum is σ = 55055000500434. Its totient is φ = 55055000500432.

The previous prime is 55055000500429. The next prime is 55055000500481. The reversal of 55055000500433 is 33400500055055.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 49095484471489 + 5959516028944 = 7006817^2 + 2441212^2 .

It is a cyclic number.

It is not a de Polignac number, because 55055000500433 - 22 = 55055000500429 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 55055000500393 and 55055000500402.

It is not a weakly prime, because it can be changed into another prime (55055000200433) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 27527500250216 + 27527500250217.

It is an arithmetic number, because the mean of its divisors is an integer number (27527500250217).

Almost surely, 255055000500433 is an apocalyptic number.

It is an amenable number.

55055000500433 is a deficient number, since it is larger than the sum of its proper divisors (1).

55055000500433 is an equidigital number, since it uses as much as digits as its factorization.

55055000500433 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 112500, while the sum is 35.

Adding to 55055000500433 its reverse (33400500055055), we get a palindrome (88455500555488).

The spelling of 55055000500433 in words is "fifty-five trillion, fifty-five billion, five hundred thousand, four hundred thirty-three".