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 Powerful Digit (Posted on 2011-03-29)
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)

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 part (1) not so hard after all (spoiler) | Comment 2 of 12 |

x and 2^x:

See correction at bottom, per Jer's later comment:

From 0 to 1, 2^x has a first digit of 1. In 1/10 of this interval, the first digit of x is also 1. So when x is between 0 and 1, the conditional probability of a first-digit match is 1/10. The overall probability contribution of this portion is therefore 1/100.

There's a lack of matches until the numbers start with 5. The initial 5's match from log(50)/log(2) (that is, the base-2 log of 50) to log(60)/log(2).

The 6's match from 6 to log(70)/log(2).

Neither the 7's nor the 8's have matches.

The 9's match from log(900)/log(2) to log(1000)/log(2).

In sum, the probability is (using notation in the Wikipedia article  of lb for the binary (base-2) logarithm):

(1/10 + lb(60) - lb(50) + lb(70) - 6 + lb(1000) - lb(900)) / 10

=

(1/10 + log(60)/log(2) - log(50)/log(2) + log(70)/log(2) - 6 + log(1000)/log(2) - log(900)/log(2)) / 10 ~= .0644320516223839

The correction per Jer's comment:

There are actually only 9 possible first digits--not 10, so the answer should be

(1/9 + lb(60) - lb(50) + lb(70) - 6 + lb(1000) - lb(900)) / 10

=

(1/9 + log(60)/log(2) - log(50)/log(2) + log(70)/log(2) - 6 + log(1000)/log(2) - log(900)/log(2)) / 10 ~= .065543162733494

Edited on March 30, 2011, 10:53 am
 Posted by Charlie on 2011-03-29 17:31:49

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