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Lucky Seven Fractions Puzzle


Date: 12/22/2001 at 16:36:11
From: Victor Hugo
Subject: Maths lucky seven problem

Put each number 1 to 9 in the X in order to make the equation correct.

   XX / XXX + XX / XX = 7


Date: 12/23/2001 at 02:48:27
From: Doctor Greenie
Subject: Re: Maths lucky seven problem

Hello, Victor -

What a great problem! I got a lot of good exercise in reasoning and 
mental math figuring this one out.

We have

    XX   XX
   --- + -- = 7
   XXX   XX

where each of the X's represents a different digit from 1 through 9.

To start with, notice that the first fraction has a 2-digit numerator 
and a 3-digit denominator, so that the first fraction is less than 1; 
with the sum of the two fractions being 7, that means the second 
fraction must be more than 6.

Then, if the fraction XX/XX, where numerator and denominator are 
2-digit numbers, is greater than 6, then the denominator must be at 
most 16. So the possible denominators are 12, 13, 14, 15, and 16.

If the denominator of the second fraction is 12, then, for the 
fraction to have a value more than 6 but less than 7, the possible 
numerators are 73 to 83. Similarly, 

  if the denominator is 13, the possible numerators are 79 to 90; 
  if the denominator is 14, the possible numerators are 85 to 97; 
  if the denominator is 15, the possible numerators are 91 to 98; and 
  if the denominator is 16, the possible numerators are 97 to 98.  

Note that the possible numerators for the last two cases are limited 
by the fact that they must be 2-digit numbers; and furthermore, of 
course, some of the possible numerators for each possible denominator 
are excluded by the requirement that all the digits in the problem be 
different.

Having narrowed the possible numerator/denominator combinations by 
rejecting those that have repeated digits, most of the remaining 
possible numerators can quickly be rejected because of the value that 
would then be required for the first fraction. For example, if the 
second fraction were 73/12, then the first fraction XX/XXX would have 
to have the value 11/12, using the digits 4, 5, 6, 8, and 9. The 
smallest possible 3-digit denominator using these 5 digits is 456, and 
you can't get a fraction whose value is 11/12 having a 2-digit 
numerator and denominator 456.

We can also reject most of the possibilities with denominator 15 (in 
fact, as it turns out, all of them) because, if the second fraction is 
XX/15, then the first fraction XX/XXX must reduce to a fraction ??/15.  
Since 0 is not one of the digits we can use, and since the digit 5 is 
used in the denominator of the second fraction, the first fraction 
must reduce to something that does not have denominator 5 or 15 - so 
it must reduce to either 1/3 or 2/3. But 1/3 = 5/15 and 2/3 = 10/15; 
then, in either case, the numerator of the first fraction would have 
to be a multiple of 5.

Following is a table of all the possibilities for the second fraction, 
with the resulting requirements for the first fraction. The numbers in 
the last column of the table refer to notes as to whether the search 
for a solution to the problem needs to be continued for the given 
second fraction. The notes are as follows:

(1) these possibilities can be rejected, because it is not possible to 
    form a first fraction, using the remaining digits, which has the 
    required value
(2) these possibilities can be rejected, because the numerator of the 
    first fraction would have to be a multiple of 5
(3) these possibilities remain to be investigated

                     then the first    and would have
                     fraction would     to be formed
    if the second      have to be        using the
     fraction is     equivalent to    remaining digits      notes
  --------------------------------------------------------------------
        73/12            11/12            45689              (1)
        74/12            10/12            35689              (1)
        75/12             9/12            34689              (1)
        76/12             8/12            34589              (1)
        78/12             6/12            34569              (1)
        79/12             5/12            34589              (1)
        83/12             1/12            45679              (3)

        79/13            12/13            24568              (1)
        82/13             9/13            45679              (1)
        84/13             7/13            25679              (1)
        85/13             6/13            24679              (1)
        86/13             5/13            24579              (3)
        87/13             4/13            24569              (3)
        89/13             2/13            24567              (3)

        85/14            13/14            23679              (1)
        86/14            12/14            23579              (1)
        87/14            11/14            23569              (1)
        89/14             9/14            23567              (1)
        92/14             6/14            35678              (1)
        93/14             5/14            25678              (1)
        95/14             3/14            23678              (3)
        96/14             2/14            23578              (3)
        97/14             1/14            23568              (3)

        92/15            13/15            34678              (1),(2)
        93/15            12/15            24678              (1),(2)
        94/15            11/15            23678              (1),(2)
        96/15             9/15            23478              (1),(2)
        97/15             8/15            23468              (1),(2)
        98/15             7/15            23467              (1),(2)

        97/16            15/16            23458              (1)
        98/16            14/16            23457              (1)

So we are left with only a very few possibilities for the second 
fraction:

   83/12; 86/13, 87/13, 89/13; 95/14, 96/14, 97/14

Note that none of these is reducible; so the fully reduced forms for 
the corresponding first fractions must be

   1/12; 5/13, 4/13, 2/13; 3/14, 2/14, 1/14

From here, the solution seems to require just a lot of work trying the 
different possibilities. My approach was to find the possible 3-digit 
denominators that could reduce to the required denominator and hope 
the remaining 2 digits can make the required numerator.

For example, with the possible second fraction of 83/12, a fraction 
equivalent to 1/12 must be made using the digits 45679. Using 
divisibility rules, a number divisible by 12 must be divisible by 3 
and by 4; so the sum of its digits must be divisible by 3 and the last 
two digits must be divisible by 4. Using the digits 45679, the 
possible denominators that satisfy those requirements are 456, 756, 
576, and 564. The fractions equivalent to 1/12 with these denominators 
are

    1    38    63    48    47
   -- = --- = --- = --- = ---
   12   456   756   576   564

All of the numerators of these possible first fractions repeat a digit 
that is either in the second fraction 83/12 or in the denominator of 
the first fraction, so there is no solution with the second fraction 
83/12.

I won't go through the details for all the other possible second 
fractions. (You might want to, since I am doing this late at night and 
have not gone back and checked all my work.) My work revealed a single 
solution to your problem:

    86    95
    -- + --- = 7
    13   247

I hope all this helps. Write back if you have any further questions or 
comments about my response. And thanks again for the nice problem!

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Logic
High School Puzzles
Middle School Logic
Middle School Puzzles

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