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.

(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