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Powerful Digit (Posted on 2011-03-29) Difficulty: 3 of 5
For a randomly chosen real number x on the interval (0,10) find the exact probability of each:

(1) That x and 2x have the same first digit

(2) That x and x2 have the same first digit

(3) That x2 and 2x have the same first digit.

(4) That x, x2 and 2x all have the same first digit.

First digit refers to the first non-zero digit of the number written in decimal form.

No Solution Yet Submitted by Jer    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts An attempt as (3), but it looks wrong according to simulation | Comment 3 of 12 |

Corrections made below, in light of Jer's correction (comment 11) are made in bold:

x^2 and 2^x:

Above 1, this isn't too hard:

2 matches 2 from sqrt(2) to log(3)/log(2), the latter aka lb(3)
3 matches 3 from sqrt(3) to 2
4 matches 4 from 2 to sqrt(5)
5 matches 5 from lb(5) to sqrt(6)
6 matches 6 from lb(6) to sqrt(7)
7 matches 7 from lb(7) to sqrt(8)
8 has probability zero of matching 8 (3^2 = 8.9999... & 2^3 = 8 -- only 1 point)
1 matches 1 from lb(10) to lb(20)
2 matches 2 again from sqrt(20) to lb(30)
9 matches 9 from lb(900) to lb(1000)

Below 1, as before, 2^x begins with a digit 1, so the question becomes What fraction of the squares begin with a digit 1?

Out of the range .1 to .99..., those x's from .1 to sqrt(2)/10 have squares beginning with a 1. Thus the probability within this range is (sqrt(2)/10 - .1) / .9.

Also, from sqrt(.1) to sqrt(.2) the squares begin with a 1. This adds to the total probability within the 0 to 1 range, making it (sqrt(2)/10 - .1 + sqrt(.2) - sqrt(.1)) / .9.

 But this same probability also applies in each of the infinitely many orders of magnitude for x, and so (sqrt(2)/10 - .1 + sqrt(.2) - sqrt(.1)) / .9 is the correct probability for all of the 0 to .999999... range.

We thus get

((sqrt(2)/10 - .1 + sqrt(.2)) / .9 +lb(3) + 2 + sqrt(5) + sqrt(6) + sqrt(7) + sqrt(8) + lb(20) +lb(30) + lb(1000) - sqrt(2) - sqrt(.1) - sqrt(3) - 2 - lb(5) - lb(6) - lb(7) - lb(10) - sqrt(20) - lb(900)) / 10

= ((sqrt(2)/10 - .1 + sqrt(.2)) / .9 +lb(3) + sqrt(5) + sqrt(6) + sqrt(7) + sqrt(8) + lb(20) +lb(30) + lb(1000) - sqrt(2) - sqrt(.1) - sqrt(3) - lb(5) - lb(6) - lb(7) - lb(10) - sqrt(20) - lb(900)) / 10

=

((sqrt(2)/10 - .1) / .9 + sqrt(.2) +log(3)/log(2) + sqrt(5) + sqrt(6) + sqrt(7) + sqrt(8) + log(20)/log(2) +log(30)/log(2) + log(1000)/log(2) - sqrt(2)  - sqrt(.1) - sqrt(3) - log(5)/log(2) - log(6)/log(2) - log(7)/log(2) - log(10)/log(2) - sqrt(20) - log(900)/log(2)) / 10

~= .264795606189634 rather than .251697023241324

This now agrees with the simulation, to a reasonable statistical difference.

I suspect this is low for some reason, as a simulation of a million trials finds 267079 matches:

DEFDBL A-Z
RANDOMIZE TIMER

PRINT (1 / 10 + LOG(60) / LOG(2) - LOG(50) / LOG(2) + LOG(70) / LOG(2) - 6 + LOG(1000) / LOG(2) - LOG(900) / LOG(2)) / 10
FOR tr = 1 TO 1000000
   x = 10 * RND(1)
   y = 2 ^ x
   s1$ = STR$(x * x): s2$ = STR$(y)
   FOR i = 1 TO LEN(s1$)
     IF INSTR("123456789", MID$(s1$, i, 1)) THEN m1$ = MID$(s1$, i, 1): EXIT FOR
   NEXT
   FOR i = 1 TO LEN(s2$)
     IF INSTR("123456789", MID$(s2$, i, 1)) THEN m2$ = MID$(s2$, i, 1): EXIT FOR
   NEXT
   IF m1$ = m2$ THEN ct = ct + 1
   PRINT ct; tr,
   PRINT USING "#.#######"; ct / tr
NEXT tr

Edited on March 30, 2011, 2:45 pm
  Posted by Charlie on 2011-03-29 22:27:42

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