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Date: Thu 2/11/99 9:49 AM
From: David L. Arnett
The answer is:
6 ? ? 6 4 5
x ? ? ? x 7 2 1
_____________ _____________
? ? ? 6 4 5
? ? ? ? 1 2 9 0
? 5 ? 5 4 5 1 5
_____________ _____________
? ? 5 ? 4 ? 4 6 5 0 4 5
I obtained this solution as follows:
First, I assigned variable names to each unknown digit in the original problem => 6ab x cde. I then went about solving each digit.
Since there are only 3 digits in the first number under the top line, it followed that e must be equal to 1, since anything larger would result in 4 digits. This yielded:
6 a b
x c d 1
_____________
6 a b
? ? ? ?
? 5 ? 5
_____________
? ? 5 ? 4 b
Since the third number under the top line ends in 5, then either c or b must be 5, and the other number (b or c) must be odd. Trying 5 for c quickly shows no way to obtain the ?5?5 for the third number under the top line, therefore b must be 5 and c must be odd. This yields:
6 a 5
x c d 1
_____________
6 a 5
? ? ? ?
? 5 ? 5
_____________
? ? 5 ? 4 5
Now to get ?5?5 for the third number under the top line, c must be either 7 or 9. c=9 would work only if a=1, but we can disprove this possibility by realizing that in order to get the 4 in the last line, a+5d must end in a 4, and no integer d will allow this. Therefore, c must be 7 by default. This yields:
6 a 5
x 7 d 1
_____________
6 a 5
? ? ? ?
4 5 ? 5
_____________
4 ? 5 ? 4 5
In order to get ?5?5 for the third number under the top line, a must be either 4 or 5. Only 4 will satisfy the previously mentioned condition that a+5d must end in a 4, so a must be 4. This yields 645 x 7d1.
6 4 5
x 7 d 1
_____________
6 4 5
? ? ? ?
4 5 1 5
_____________
4 ? 5 ? 4 5
For the final number (d), we realize that in order for the final number to have a 4 for its second to last digit, d must be even. Plugging in numbers identifies 2 as the correct solution, yielding:
6 4 5
x 7 2 1
_____________
6 4 5
1 2 9 0
4 5 1 5
_____________
4 6 5 0 4 5
It is, of course, debatable whether it was quicker to reason out an answer in this fashion or to write a program to search out all possibilities for the original 5 ?'s above the first line, since there would be only 100,000 variations to try.
Dave Arnett
Copyright © 1996-2008 Alexander Bogomolny
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