Three FractionsDate: 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 proof.) 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 a -- 1c 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 here 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 http://mathforum.org/dr.math/ 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. |
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