Lucky Seven Fractions PuzzleDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/