Base | Representation |
---|---|
bin | 111110011101000101… |
… | …1101101000110101011 |
3 | 110211012001200200121122 |
4 | 1330322023231012223 |
5 | 4144134223224120 |
6 | 141340335525455 |
7 | 12455423215010 |
oct | 1747213550653 |
9 | 424161620548 |
10 | 134120133035 |
11 | 51975356059 |
12 | 21bb068988b |
13 | c855668885 |
14 | 66c4771707 |
15 | 374e918a25 |
hex | 1f3a2ed1ab |
134120133035 has 16 divisors (see below), whose sum is σ = 183970200000. Its totient is φ = 91951082496.
The previous prime is 134120132999. The next prime is 134120133071. The reversal of 134120133035 is 530331021431.
It is an interprime number because it is at equal distance from previous prime (134120132999) and next prime (134120133071).
It is not a de Polignac number, because 134120133035 - 210 = 134120132011 is a prime.
It is a super-2 number, since 2×1341201330352 (a number of 23 digits) contains 22 as substring.
It is a junction number, because it is equal to n+sod(n) for n = 134120132995 and 134120133013.
It is an unprimeable number.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 160910 + ... + 542339.
It is an arithmetic number, because the mean of its divisors is an integer number (11498137500).
Almost surely, 2134120133035 is an apocalyptic number.
134120133035 is a deficient number, since it is larger than the sum of its proper divisors (49850066965).
134120133035 is an equidigital number, since it uses as much as digits as its factorization.
134120133035 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 708710.
The product of its (nonzero) digits is 3240, while the sum is 26.
Adding to 134120133035 its reverse (530331021431), we get a palindrome (664451154466).
The spelling of 134120133035 in words is "one hundred thirty-four billion, one hundred twenty million, one hundred thirty-three thousand, thirty-five".
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