Base | Representation |
---|---|
bin | 11111000101111000110011… |
… | …00110001001000101001111 |
3 | 22222002022010012101202001110 |
4 | 33202330121212021011033 |
5 | 32430201341014022340 |
6 | 401225400341343103 |
7 | 20254500145255155 |
oct | 1742743146110517 |
9 | 288068105352043 |
10 | 68372013814095 |
11 | 1a870454385a82 |
12 | 7802b55107a93 |
13 | 2c1c5b79bb864 |
14 | 12c531a3833d5 |
15 | 7d87a64c8680 |
hex | 3e2f1998914f |
68372013814095 has 16 divisors (see below), whose sum is σ = 115830235167840. Its totient is φ = 34320069679104.
The previous prime is 68372013814081. The next prime is 68372013814133. The reversal of 68372013814095 is 59041831027386.
It is not a de Polignac number, because 68372013814095 - 210 = 68372013813071 is a prime.
It is a congruent number.
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, 134062771930 + ... + 134062772439.
It is an arithmetic number, because the mean of its divisors is an integer number (7239389697990).
It is a 1-persistent number, because it is pandigital, but 2⋅68372013814095 = 136744027628190 is not.
Almost surely, 268372013814095 is an apocalyptic number.
68372013814095 is a deficient number, since it is larger than the sum of its proper divisors (47458221353745).
68372013814095 is a wasteful number, since it uses less digits than its factorization.
68372013814095 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 268125544394.
The product of its (nonzero) digits is 8709120, while the sum is 57.
The spelling of 68372013814095 in words is "sixty-eight trillion, three hundred seventy-two billion, thirteen million, eight hundred fourteen thousand, ninety-five".
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