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Lucky Seven Fractions Problem

Date: 06/08/2002 at 00:02:44
From: LJ Lanni
Subject: fractions

??/??? + ??/?? = 7

? = a number from 1 to 9 (not 0)

Put each of the numbers 1 through 9 in place of a ? mark in order 
to make the equation correct.


Date: 06/19/2002 at 01:29:19
From: Doctor Greenie
Subject: Re: fractions

Hello LJ,

Let me rewrite your equation as

    AB   FG
   --- + -- = 7
   CDE   HI

where the capital letters are the digits 1 through 9 (i.e., "AB" is a 
2-digit number; it does not imply multiplication of the "A" and "B").

Then...

(1) The first fraction is less than 1, because the numerator is two 
digits and the denominator is three digits

(2) If the sum of the two fractions is 7 and the first fraction is 
less than 1, then the second fraction is equivalent to a mixed number 
between 6 and 7

(3) For the second fraction to be a mixed number between 6 and 7, the 
digit "H" must be 1

(4) Since "H" is 1, "C" must be at least 2; this means that the first 
fraction is less than 1/2, and so the second fraction must be 
equivalent to a mixed number between 6.5 and 7

First possibility:  I=2

Let's first consider the possibility that I=2, so C is 3 or greater.  
Then the first fraction is less than 1/3 and so the second fraction 
must be greater than 6 2/3 (and of course less than 7):

           FG
   6 2/3 < -- < 7 
           12

or

   80 < FG < 84

Also, we have used the digits 1 and 2 already, so our only 
possibility here is that FG is 83.  Then we have

   AB   83
  --- + -- = 7
  CDE   12

so

   AB    1
  --- = --
  CDE   12

The digits we have left for ABCDE are {4, 5, 6, 7, 9}.  With these 
digits, and with the fraction reducing to 1/12, the only 
possibilities for the fraction AB/CDE are with B=7 and E=4:

   AB    ?7
  --- = ---
  CDE   ??4

But no combination of the digits 5, 6, and 9 in the remaining places 
results in a fraction equivalent to 1/12.

So there is no solution with I=2.

Second possibility:  I=3

If I=3, then C can be 2; let's investigate this combination next.  We 
have

   AB    FG
  --- +  -- = 7
  2DE    13

The first fraction is less than 1/2, so the second fraction is 
greater than 6 1/2 and of course less than 7:

   13   FG
   -- < -- < 7
    2   13

which leads us to

   84.5 < FG < 91

So we need to consider the following possible values of FG: 85, 86, 
87, and 89 (88 is not allowed because F and G are different; 90 is 
not allowed because 0 is not one of the digits we can use).  So we 
see that F must be 8.  And we now have

   AB   8G
  --- + -- = 7
  2DE   13

For the sum of these two fractions to be equal to the whole number 7, 
the number "2DE" must be a multiple of 13, so that the 
fraction "AB/2DE" will reduce to a fraction having a denominator of 
13.  

The multiples of 13 between 200 and 299 are the following: 208, 
221, 234, 247, 260, 273, 286, and 299.  Since we have used the digits 
1, 2, 3, and 8, and since repeated digits are not allowed, and since 
we can't use 0, the only possibility remaining from that list of 
multiples of 13 is 247.

We now have

   AB   8G      AB      8G
  --- + -- = -------- + --
  247   13   (19)(13)   13

The remaining digits for ABG are 5, 6, and 9; and the number "AB" 
must be a multiple of 19.  95 is a multiple of 19; that leaves G=6.

Checking out this possibility, we find that it is indeed a solution 
to the problem:

   95   86   (5)(19)    86    5   86   91
  --- + -- = -------- + -- = -- + -- = -- = 7
  247   13   (13)(19)   13   13   13   13

I sketched out an analysis for the case where I=3 and C>3; and 
started on the sketch of the analysis for the case with I=4 and C=2, 
but, having found one solution to the problem, my interest flagged....
but the point is that there may be other solutions to the problem.

I hope your curiosity is satisfied now.  Thanks for the mental 
exercise your problem provided me with.

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
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