The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Three Fractions

Date: 09/06/2001 at 08:53:40
From: Kathleen Hazel Reyes
Subject: Fractions

Three Fractions

a/bc + d/ef + g/hi = 1

Each of three fractions has a one-digit numerator and a two-digit 
denominator. The three fractions together add up to one. Place the 
nine digits 1-9 into the fractions to make the equation correct.

Source: Nob Yoshigara, c/o Scot Morris in OMNI, April 1994. 

I can't proceed because there so many possibilities in jumbling those 
numbers. I tried 2/34 + 5/68 + 7/91 but the sum of this is not equal 
to one. Please help me!

Date: 09/06/2001 at 14:22:24
From: Doctor Greenie
Subject: Re: Fractions

Hi, Kathleen --

This is a nice problem for teaching reasoning skills as well as 
arithmetic skills.

I found one solution analytically and did some "gut feeling" reasoning 
to decide there were no others; then I wrote a short computer program 
that confirmed the single solution. The response below describes my 
analysis of the problem.

You show as one example of the things you tried, the following:

     2    5    7
    -- + -- + --
    34   68   91

Let's look at this trial answer and consider some of the problems it 
has. For one thing, each of these fractions is less than 1/10, so the 
sum won't be anywhere near 1. It appears that, if the sum of the three 
fractions is to be 1, then at least one of the fractions is going to 
have to have a "small" denominator and a "large" numerator.

A "small" denominator will probably mean the 2-digit number "1x".  
Let's see if we can prove (informally, at least) that one of the 
denominators must be "1x".  To do this, let's assume that this is NOT 
the case - that the denominators are as "small" as they can be without 
one of them being "1x". Then our sum looks like:

     x    x    x
    -- + -- + --
    2x   3x   4x

Now let's try the combination of digits in this pattern that gives us 
the largest numerators and the smallest denominators:

     9    8    7
    -- + -- + --
    21   35   46

This sum is far short of 1, so it looks as if one of the denominators 
must be "1x".  (I say "looks as if" because we have not done a formal 

So it seems our solution is going to have to be of the following form:

     x    x    x
    -- + -- + --
    1x   xx   xx

Let's now go back and look again at your trial solution and see 
another problem it has.

     2    5    7
    -- + -- + --
    34   68   91

The denominators "34" and "68" look promising, because 34*2 = 68, so 
those two fractions will combine nicely. The third fraction, 7/91, 
looks a bit promising at first, because 7/91 = 1/13.  But now we are 
in real trouble, because 13 is a prime number, and the fraction 1/13 
will not combine nicely with any other fractions except ones in which 
the denominator is a multiple of 13. Unfortunately, 34 and 68 are not 
multiples of 13, so there is no chance that the sum of the three 
fractions will equal 1.

So we need to have denominators that combine "nicely."

We have decided that:

(a) the first fraction should be a/1c, where a is "large"; and
(b) the denominators need to combine "nicely."

From here, the solution of the problem becomes a trial-and-error 
process, outlined as follows:

(1) Choose a first fraction of the form


For the sum of the three fractions to be 1, the numerator a of this 
fraction should be "large" (and c should be "small").

(2) Given the first fraction you have chosen, determine what the sum 
of the other two fractions must be in order to make the sum of all 
three fractions equal to 1.

(3) With the 6 digits that are left, see if you can find two 
denominators that might combine "nicely" to give you the sum you need 
for the remaining two fractions.

(4) With these guesses for the denominators of the remaining two 
fractions, there are only 2 digits left for the numerators of those 
fractions. See if either combination of those two remaining digits as 
the numerators in the fractions with these guesses for the 
denominators gives a solution to the problem.

Note that, in the first step, if we choose 9/12 for our first 
fraction, this reduces to 3/4, so the remaining two fractions must 
add to 1/4. If, on the other hand, we choose 9/13 for our first 
fraction, then the remaining two fractions must add to 4/13.  We are 
much more likely to find two 2-digit denominators formed from the 
digits 3-4-5-6-7-8, which can combine to give a sum of 1/4, than we 
are to find two 2-digit denominators formed from the digits 
2-4-5-6-7-8, which can combine to give a sum of 4/13.  So we might 
concentrate our search on first fractions that reduce to lower terms.

Following is the beginning of my search for solutions using this 
trial-and-error process:

 1) try first fraction = 9/12 = 3/4
 2) the sum of the remaining two fractions must be 1/4
 3) with the digits 3-4-5-6-7-8 remaining, I see only two promising  
    pairs of 2-digit denominators that might give us fractions whose 
    sum is 1/4 - 36 and 48, or 34 and 68
4a) try denominators 36 and 48; 7/36 + 5/48 = 41/144 is not equal to 
    1/4, and 5/36 + 7/48 = 43/144 is not equal to 1/4, so no solution 
4b) try denominators 34 and 68; 7/34 + 5/68 = 19/68 is not equal to 
    1/4, but 5/34 + 7/68 = 17/68 IS equal to 1/4

We have a solution using 9/12 as our first fraction.

     9    5    7
    -- + -- + -- = 1
    12   34   68

But let's look for others....

 1) try first fraction = 8/12 = 2/3
 2) the sum of the remaining two fractions must be 1/3
 3) with the digits 3-4-5-6-7-9 remaining, and with the sum of the 
    remaining two fraction needing to be 1/3, I see four promising 
    denominators - 36, 45, 54, and 63. With each digit being used only 
    once, the possible pairs of denominators are 36 and 45, 36 and 54, 
    45 and 63, or 54 and 63; the numerators are 7 and 9
 4) All the promising denominators are multiples of 9, and 9 is one of 
    the two denominators. The fraction with 9 in the numerator will 
    reduce to lower terms in every case; the fraction with 7 in the 
    numerator will not reduce except in the case with denominator 63, 
    and that combination does not lead to a solution

There are no solutions using 8/12 as our first fraction.

I didn't go through the details with 7/12 as a guess for my first 
fraction; I felt it was too "small" for the largest of my fractions to 
allow a sum of 1.

 1) try first fraction = 9/13
 2) the sum of the remaining two fractions must be 4/13
 3) with the digits 2-4-5-6-7-8 remaining, and with the sum of the 
    remaining two fraction needing to be 4/13, the only promising 
    denominators are those that are multiples of 13: 26, 52, 65, and 
    78. The possible combinations of these denominators, with the 
    restriction that each digit is only used once, are 26 and 78, 52 
    and 78, and 65 and 78.
 4) With any of these possible combinations of denominators, the two 
    digits remaining are too small to have the sum of these last two 
    fractions equal to the required 4/13

There are no solutions using 9/13 as our first fraction.

I did not pursue any other possible solutions using different guesses 
for a first fraction; the work required to check all the possibilities 
for a particular first fraction is tedious, and I felt that the other 
possible guesses for the first fraction did not leave reasonable 
chances for a solution to the problem.

You submitted this same question twice, worded differently.  In your 
other message, you state that your professor claims there are two 
solutions. If you don't consider adding the same three fractions in a 
different order as a "different" solution, then I disagree.  I wrote a 
computer program that checked ALL possible arrangements of the nine 
digits 1-9 in the three fractions, and the only solution the program 
found was the one we found above.

Write back if you have any further questions on this (or if there 
really is a second distinct solution).

- Doctor Greenie, The Math Forum   

Date: 09/07/2001 at 20:39:48
From: Kathleen Hazel Reyes
Subject: Re: Fractions

Thanks really for your interest in my problem. Lots of thanks.
Associated Topics:
High School Puzzles

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.