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13353443450497 is a prime number
BaseRepresentation
bin1100001001010001011101…
…0000001101111010000001
31202021120120102201010201001
43002110113100031322001
53222240323010403442
644222253124212001
72545516530353053
oct302242720157201
952246512633631
1013353443450497
1142891887823a7
1215b7ba123a001
1375b2bba83ac7
1434244ad358d3
15182548b772b7
hexc251740de81

13353443450497 has 2 divisors, whose sum is σ = 13353443450498. Its totient is φ = 13353443450496.

The previous prime is 13353443450407. The next prime is 13353443450551. The reversal of 13353443450497 is 79405434435331.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 9775383686721 + 3578059763776 = 3126561^2 + 1891576^2 .

It is a cyclic number.

It is not a de Polignac number, because 13353443450497 - 239 = 12803687636609 is a prime.

It is a super-3 number, since 3×133534434504973 (a number of 40 digits) contains 333 as substring.

It is not a weakly prime, because it can be changed into another prime (13353443450407) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6676721725248 + 6676721725249.

It is an arithmetic number, because the mean of its divisors is an integer number (6676721725249).

Almost surely, 213353443450497 is an apocalyptic number.

It is an amenable number.

13353443450497 is a deficient number, since it is larger than the sum of its proper divisors (1).

13353443450497 is an equidigital number, since it uses as much as digits as its factorization.

13353443450497 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 32659200, while the sum is 55.

The spelling of 13353443450497 in words is "thirteen trillion, three hundred fifty-three billion, four hundred forty-three million, four hundred fifty thousand, four hundred ninety-seven".