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Hexadecimal Hinder (Posted on 2016-06-19) Difficulty: 3 of 5
x is a randomly chosen hexadecimal real number on the interval (0, (10)16)

Determine the probability of each of the following:

(i) x and 3x have the same first digit.

(ii) x and x3 have the same first digit.

(iii) x3 and 3x have the same first digit.

(iv) x, x3 and 3x have the same first digit.

*** “First digit” denotes the first nonzero digit of the number when expressed in hexadecimal form.

No Solution Yet Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

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Solution part (ii) solution | Comment 2 of 6 |
(ii)
The probability should be the same within any span of numbers starting with 1 and ending with 10 (i.e., 10 hex), such as 1 - 10 or .1 to 1 or .01 to .1, etc., as the digits of the cubes depend only on the digits of x.

So the probability within (0,10 hex) is the same as within (1,10 hex), a span of 15 (in decimal).

 1                      1                                                       
 2                      8                                                       
 3                      1B                                                      
 4                      40                                                      
 5                      7D                                                      
 6                      D8                                                      
 7                      157                                                     
 8                      200                                                     
 9                      2D9                                                     
 A                      3E8                                                     
 B                      533                                                     
 C                      6C0                                                     
 D                      895                                                     
 E                      AB8                                                     
 F                      D2F                                                     
 10                     1000                                                    

In the first range, 1 through cuberoot(2) qualify, so

2^(1/3) - 1 = .25992104989487, which is over a quarter of a unit on the number line,

which becomes the first addend in the sum that will be divided by 15.

When 2 plus a fraction is cubed, the result begins with 8 through F, or 1 -- never a 2.

When 3 plus a fraction is cubed, the result begins with 1 through 3:

(4*16)^(1/3) - (3*16)^(1/3) = .36575881433572, the second addend

When 4 plus a fraction is cubed, the result begins with 4 through 7:

(5*16)^(1/3) - (4*16)^(1/3)  = .30886938006377, the third addend

5 plus a fraction results in from 7 through D-- nothing to count here.

In fact nothing else has matches until F:

(16^3)^(1/3) - (15*16^2)^(1/3) = .3405294353246, the last addend.

.25992104989487
.36575881433572
.30886938006377
.3405294353246

totalling   1.27507867961896. When divided by 15, it gives us

 .08500524530793067 as the probability.

  Posted by Charlie on 2016-06-19 18:44:57
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