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