Base | Representation |
---|---|
bin | 11110011110100101100101… |
… | …001010010110011101010111 |
3 | 122120121102021020122221220102 |
4 | 132132211211022112131113 |
5 | 120032131220334144421 |
6 | 1153030353441324315 |
7 | 40143211466635331 |
oct | 3636454512263527 |
9 | 576542236587812 |
10 | 134043331553111 |
11 | 3978a504183012 |
12 | 1304a5b463669b |
13 | 59a4302712376 |
14 | 2515a3633a451 |
15 | 1076b9477260b |
hex | 79e965296757 |
134043331553111 has 16 divisors (see below), whose sum is σ = 138807765720000. Its totient is φ = 129288444612096.
The previous prime is 134043331553099. The next prime is 134043331553141. The reversal of 134043331553111 is 111355133340431.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-134043331553111 is a prime.
It is a super-2 number, since 2×1340433315531112 (a number of 29 digits) contains 22 as substring.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (134043331553141) by changing a digit.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 117387305 + ... + 118523693.
It is an arithmetic number, because the mean of its divisors is an integer number (8675485357500).
Almost surely, 2134043331553111 is an apocalyptic number.
134043331553111 is a deficient number, since it is larger than the sum of its proper divisors (4764434166889).
134043331553111 is a wasteful number, since it uses less digits than its factorization.
134043331553111 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 1140586.
The product of its (nonzero) digits is 97200, while the sum is 38.
Adding to 134043331553111 its reverse (111355133340431), we get a palindrome (245398464893542).
The spelling of 134043331553111 in words is "one hundred thirty-four trillion, forty-three billion, three hundred thirty-one million, five hundred fifty-three thousand, one hundred eleven".
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