Feynman Long Division

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Feynman Long Division

Postby Lanzaa » Fri Apr 06, 2007 6:36 am UTC

I searched the forum for long division and did not find anything.

http://en.wikipedia.org/wiki/Feynman_Long_Division_Puzzles

The following represents a long division. Each dot represents any digit, and all the "A"'s represent the same digit. None of the the dots are the same as the "A" digit. Find all the digits in the long division algorithm.

Code: Select all
   ____..A.
.A.\....A..
    ..AA
    -----
     ...A
      ..A
      -----
      ....
      .A..
      ------
       ....
       ....
      ------
          0


It took me 70 minutes to solve. The challenge is to solve it without a calculator.
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Postby hyperion » Fri Apr 06, 2007 9:41 am UTC

I can't remember how to do long division. Now I just guess, by already knowing most 'division rules'. Stuff like 37/7 would be around 5.3.

Carry on!
Peshmerga wrote:A blow job would probably get you a LOT of cheeseburgers.
But I digress.
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Postby Drostie » Fri Apr 06, 2007 10:16 am UTC

Took me around the same time--60 minutes, ish.

Big hint for anyone who's totally stuck:

The Hints wrote:Your first task is to determine A. Write out the 9x9 times table, then write out the relations:

[1] (a)*(xAy) = (__A)
[2] (b)*(xAy) = (__AA)
[3] (A)*(xAy) = (_A__)

Now, [3] guarantees that A is not 1. For every other number 2 through 9, try to fit [1] and [2] on the times table. So, let's say I'm looking at A=4. Now I need to look through the times table to find one column that has two different numbers, a and b, such that a*y has last digit 4, and b*y has last digit 4. Since [2] is larger than [1], you know (b) is larger than (a).

Four columns have this: 2, 4, 6, and 8. Column 4 can't be right, because A is 4 and y is not A.

For the other three columns, 2, 6, and 8, do the multiplications on equation [2] to look for whether they get that second digit right.

For example, 7*(x42) = __94 gets the second digit wrong, so now I know that if A=4, then y could not be 2. That eliminates the 2-column. Repeat for y=6 and y=8, and then repeat the process for all other numbers A could be, to find a small set of numbers which A could be.


When you're done:
The Answer wrote:3527876 divided by 484 equals 7289.
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Postby Token » Fri Apr 06, 2007 4:38 pm UTC

It's not as bad as it seems, if you just approach it methodically. Took me about 15 minutes.

Solution wrote:
Looking at the first line, we have .A. * . = ..AA

If we take only the last two digits, we can search for As that fit AB * C = DAA, where A,B,C,D are distinct. We can see that A is not 0 or 5, and a brute force check of possibilities (not too bad, since D is limited) gives us three options:

37 x 9 = 333
43 x 8 = 344
84 x 7 = 588

Then, looking at the second line, we have .AB * . = ..A

Now, looking at our possibilities, we need a single digit E such that BxE ends in D. The only possibility we have where E is not C is the third, where E could be 2. In fact, this option is the only one that works, since the others give answers that exceed three digits.

Now we can look at the second line, where we are helpfully given that the single digit multiplier is 8 (since we now know A=8). So we have .84 * 8 = .8..

The last two digits are easy to work out, so we get F84 x 8 = G872. Subtract 84*8 from each side, so F00 * 8 = G200. F is equal to 4 or 9, but 9 does not fit the second line of multiplication. So we have F = 4.

A bit of multiplication now gives us that we are looking for 484 x 352.8.. = 728H

The only H that gives us the 8 where we want it is H=9. So we have the final answer, that 484 x 7289 = 3527876.
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