Base | Representation |
---|---|
bin | 100110101010101100010… |
… | …1100100110111101111111 |
3 | 200211001022001212120012211 |
4 | 1031111120230212331333 |
5 | 1144032303313242421 |
6 | 15145215014441251 |
7 | 1055643601030366 |
oct | 115253054467577 |
9 | 20731261776184 |
10 | 5314362634111 |
11 | 1769899186403 |
12 | 719b61363227 |
13 | 2c71b068c72b |
14 | 1453061da7dd |
15 | 9338ac959e1 |
hex | 4d558b26f7f |
5314362634111 has 2 divisors, whose sum is σ = 5314362634112. Its totient is φ = 5314362634110.
The previous prime is 5314362634079. The next prime is 5314362634219. The reversal of 5314362634111 is 1114362634135.
5314362634111 is digitally balanced in base 3, because in such base it contains all the possibile digits an equal number of times.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 5314362634111 - 25 = 5314362634079 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (5314369634111) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2657181317055 + 2657181317056.
It is an arithmetic number, because the mean of its divisors is an integer number (2657181317056).
Almost surely, 25314362634111 is an apocalyptic number.
5314362634111 is a deficient number, since it is larger than the sum of its proper divisors (1).
5314362634111 is an equidigital number, since it uses as much as digits as its factorization.
5314362634111 is an evil number, because the sum of its binary digits is even.
The product of its digits is 155520, while the sum is 40.
The spelling of 5314362634111 in words is "five trillion, three hundred fourteen billion, three hundred sixty-two million, six hundred thirty-four thousand, one hundred eleven".
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