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13320211134343 is a prime number
BaseRepresentation
bin1100000111010101101001…
…1100111111011110000111
31202011101210102020120010211
43001311122130333132013
53221214243022244333
644155115344300251
72543232151341001
oct301653234773607
952141712216124
1013320211134343
114276084a23403
1215b1667911687
13758123b46bc9
143409b74cd371
151817513b9ecd
hexc1d5a73f787

13320211134343 has 2 divisors, whose sum is σ = 13320211134344. Its totient is φ = 13320211134342.

The previous prime is 13320211134313. The next prime is 13320211134383. The reversal of 13320211134343 is 34343111202331.

It is a weak prime.

It is an emirp because it is prime and its reverse (34343111202331) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 13320211134343 - 213 = 13320211126151 is a prime.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (13320211134313) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6660105567171 + 6660105567172.

It is an arithmetic number, because the mean of its divisors is an integer number (6660105567172).

Almost surely, 213320211134343 is an apocalyptic number.

13320211134343 is a deficient number, since it is larger than the sum of its proper divisors (1).

13320211134343 is an equidigital number, since it uses as much as digits as its factorization.

13320211134343 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 15552, while the sum is 31.

Adding to 13320211134343 its reverse (34343111202331), we get a palindrome (47663322336674).

The spelling of 13320211134343 in words is "thirteen trillion, three hundred twenty billion, two hundred eleven million, one hundred thirty-four thousand, three hundred forty-three".