Base | Representation |
---|---|
bin | 1100010010111111000011… |
… | …1100010001001100100111 |
3 | 1202212112021021001012202012 |
4 | 3010233300330101030213 |
5 | 3233004040412131211 |
6 | 44431050405504435 |
7 | 2563544514206606 |
oct | 304576074211447 |
9 | 52775237035665 |
10 | 13520304411431 |
11 | 4342a24404004 |
12 | 16243aa71211b |
13 | 770c6159480c |
14 | 34a559b31a3d |
15 | 186a62b16d8b |
hex | c4bf0f11327 |
13520304411431 has 2 divisors, whose sum is σ = 13520304411432. Its totient is φ = 13520304411430.
The previous prime is 13520304411403. The next prime is 13520304411443. The reversal of 13520304411431 is 13411440302531.
13520304411431 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 13520304411431 - 214 = 13520304395047 is a prime.
It is a super-2 number, since 2×135203044114312 (a number of 27 digits) contains 22 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (13520304411631) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6760152205715 + 6760152205716.
It is an arithmetic number, because the mean of its divisors is an integer number (6760152205716).
Almost surely, 213520304411431 is an apocalyptic number.
13520304411431 is a deficient number, since it is larger than the sum of its proper divisors (1).
13520304411431 is an equidigital number, since it uses as much as digits as its factorization.
13520304411431 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 17280, while the sum is 32.
Adding to 13520304411431 its reverse (13411440302531), we get a palindrome (26931744713962).
The spelling of 13520304411431 in words is "thirteen trillion, five hundred twenty billion, three hundred four million, four hundred eleven thousand, four hundred thirty-one".
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