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323201156480497 is a prime number
BaseRepresentation
bin100100101111100110010001…
…0000111110111110111110001
31120101100201200022002100110101
41021133030202013313313301
5314330311432024333442
63103220321143232401
7125035340644604251
oct11137144207676761
91511321608070411
10323201156480497
1193a88927aa6339
12302ba671448701
1310b4597a0025a9
1459b503b4ad961
15275730cb9dcb7
hex125f3221f7df1

323201156480497 has 2 divisors, whose sum is σ = 323201156480498. Its totient is φ = 323201156480496.

The previous prime is 323201156480437. The next prime is 323201156480507. The reversal of 323201156480497 is 794084651102323.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 202138756437601 + 121062400042896 = 14217551^2 + 11002836^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-323201156480497 is a prime.

It is not a weakly prime, because it can be changed into another prime (323201156480417) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 161600578240248 + 161600578240249.

It is an arithmetic number, because the mean of its divisors is an integer number (161600578240249).

It is a 1-persistent number, because it is pandigital, but 2⋅323201156480497 = 646402312960994 is not.

Almost surely, 2323201156480497 is an apocalyptic number.

It is an amenable number.

323201156480497 is a deficient number, since it is larger than the sum of its proper divisors (1).

323201156480497 is an equidigital number, since it uses as much as digits as its factorization.

323201156480497 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 8709120, while the sum is 55.

The spelling of 323201156480497 in words is "three hundred twenty-three trillion, two hundred one billion, one hundred fifty-six million, four hundred eighty thousand, four hundred ninety-seven".