Base | Representation |
---|---|
bin | 11110011110000001100010… |
… | …100000110101110001000111 |
3 | 122120110201101020101212020102 |
4 | 132132001202200311301013 |
5 | 120031012441344124312 |
6 | 1153000513412354315 |
7 | 40140341466440615 |
oct | 3636014240656107 |
9 | 576421336355212 |
10 | 134004632411207 |
11 | 39775055558a08 |
12 | 13042bb431599b |
13 | 59a077600852c |
14 | 2513c048346b5 |
15 | 1075b7c099ec2 |
hex | 79e062835c47 |
134004632411207 has 2 divisors, whose sum is σ = 134004632411208. Its totient is φ = 134004632411206.
The previous prime is 134004632411189. The next prime is 134004632411401. The reversal of 134004632411207 is 702114236400431.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 134004632411207 - 212 = 134004632407111 is a prime.
It is a super-5 number, since 5×1340046324112075 (a number of 72 digits) contains 55555 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (134004632411107) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 67002316205603 + 67002316205604.
It is an arithmetic number, because the mean of its divisors is an integer number (67002316205604).
Almost surely, 2134004632411207 is an apocalyptic number.
134004632411207 is a deficient number, since it is larger than the sum of its proper divisors (1).
134004632411207 is an equidigital number, since it uses as much as digits as its factorization.
134004632411207 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 96768, while the sum is 38.
Adding to 134004632411207 its reverse (702114236400431), we get a palindrome (836118868811638).
The spelling of 134004632411207 in words is "one hundred thirty-four trillion, four billion, six hundred thirty-two million, four hundred eleven thousand, two hundred seven".
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