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13552106035301 is a prime number
BaseRepresentation
bin1100010100110101100001…
…1101110101100001100101
31202222120100120102002110022
43011031120131311201211
53234014203121112201
644453430201004525
72566051551435311
oct305153035654145
952876316362408
1013552106035301
114355463610789
12162a5a4a83745
13773c5bc97b32
1434bcd5556b41
151877c49ba41b
hexc5358775865

13552106035301 has 2 divisors, whose sum is σ = 13552106035302. Its totient is φ = 13552106035300.

The previous prime is 13552106035297. The next prime is 13552106035333. The reversal of 13552106035301 is 10353060125531.

13552106035301 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 13492765562500 + 59340472801 = 3673250^2 + 243599^2 .

It is a cyclic number.

It is not a de Polignac number, because 13552106035301 - 22 = 13552106035297 is a prime.

It is a super-3 number, since 3×135521060353013 (a number of 40 digits) contains 333 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (13552106035381) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6776053017650 + 6776053017651.

It is an arithmetic number, because the mean of its divisors is an integer number (6776053017651).

Almost surely, 213552106035301 is an apocalyptic number.

It is an amenable number.

13552106035301 is a deficient number, since it is larger than the sum of its proper divisors (1).

13552106035301 is an equidigital number, since it uses as much as digits as its factorization.

13552106035301 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 40500, while the sum is 35.

The spelling of 13552106035301 in words is "thirteen trillion, five hundred fifty-two billion, one hundred six million, thirty-five thousand, three hundred one".