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Hexadecimal Digit Travail (Posted on 2014-12-25) Difficulty: 3 of 5
Determine the probability that a for a seven-digit positive hexadecimal integer N, the sum of the first four digits of N is equal to the product of the last three digits of N.

No Solution Yet Submitted by K Sengupta    
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re: computer solution | Comment 2 of 5 |
(In reply to computer solution by Charlie)

Having seen your result, finding it close but not the same as mine, I am not sure the procedure I followed is correct. Perhaps you can tell me where I erred, if I had.

I don't have the processing speed that I would guess you do,
so it would take much, much longer than 1/2 hour run time to have computed it in same manner as you had, though it took only about that long for the result that I came up with.

I divided the 7-digit number into two parts, the first four digits and the last three digits. As the product of the last three digits will be between 0 and 4096 decimal, I computationally counted the number of times a distinct product appeared
between between 000 and FFF. The probability for each of products was then calculated as the count/4096. As the sum of the first 4 digits will be between 1 and 60, I computationally counted the number of times a distinct sum appeared between 1000 and FFFF, and then divided each count by 61440. Then for each of the distinct numbers between 1 and 60, I multiplied the probability of the product and the probability of the sum, (the products that are larger than 60 would have 0 probability of equalling the sum, thus did not need to be calculated), then I took the sum of these products which were calculated as approximately 0.002059412.  The result is only  0.00000701 larger than your calculated result, which is not a great amount, but one that causes me to question my method.

  Posted by Dej Mar on 2014-12-25 17:16:18

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