Difference between revisions of "1966 AHSME Problems/Problem 39"
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<math>F_1=\frac{3R_1+7}{R_1^2-1}=\frac{2R_2+5}{R_2^2-1}</math> and <math>F_2=\frac{7R_1+3}{R_1^2-1}=\frac{5R_2+2}{R_2^2-1}</math>. | <math>F_1=\frac{3R_1+7}{R_1^2-1}=\frac{2R_2+5}{R_2^2-1}</math> and <math>F_2=\frac{7R_1+3}{R_1^2-1}=\frac{5R_2+2}{R_2^2-1}</math>. | ||
Dividing the 2 equations to get some potential solutions, we get (note that we don't have to worry about division by zero because <math>R_1,R_2>7</math>): | Dividing the 2 equations to get some potential solutions, we get (note that we don't have to worry about division by zero because <math>R_1,R_2>7</math>): | ||
| − | <math>\frac{3F_1+7}{7F_1+3}=\frac{2F_2+5}{5F_2+2}\\15F_1F_2+6F_1+35F_2+14=14F_1F_2+35F_1+6F_2+15\\F_1F_2-29F_1+29F_2=1\\\left(F_1+29\right)\left( | + | <math>\frac{3F_1+7}{7F_1+3}=\frac{2F_2+5}{5F_2+2}\\15F_1F_2+6F_1+35F_2+14=14F_1F_2+35F_1+6F_2+15\\F_1F_2-29F_1+29F_2=1\\\left(F_1+29\right)\left(F_2-29\right)=-840</math> |
Since non-integer bases are rarely used, we can try to assume that the solution is positive integers and see if we get any solutions. | Since non-integer bases are rarely used, we can try to assume that the solution is positive integers and see if we get any solutions. | ||
Notice that <math>(F_2-29)</math> has to be a negative factor of 840. We need to plug in values of <math>F_2 > 7</math>. 840 divides -21, so we plug in 8 to check. Luckily, when <math>F_2 = 8</math>, we see that <math>F_1=11</math>, and furthermore, 11+8=19 is one of the answers. We can quickly test it into the original equations <math>F_1=\frac{3R_1+7}{R_1^2-1}=\frac{2R_2+5}{R_2^2-1}</math> and <math>F_2=\frac{7R_1+3}{R_1^2-1}=\frac{5R_2+2}{R_2^2-1}</math>, and we see they indeed work. Therefore the answer is <math>\boxed{E}</math> | Notice that <math>(F_2-29)</math> has to be a negative factor of 840. We need to plug in values of <math>F_2 > 7</math>. 840 divides -21, so we plug in 8 to check. Luckily, when <math>F_2 = 8</math>, we see that <math>F_1=11</math>, and furthermore, 11+8=19 is one of the answers. We can quickly test it into the original equations <math>F_1=\frac{3R_1+7}{R_1^2-1}=\frac{2R_2+5}{R_2^2-1}</math> and <math>F_2=\frac{7R_1+3}{R_1^2-1}=\frac{5R_2+2}{R_2^2-1}</math>, and we see they indeed work. Therefore the answer is <math>\boxed{E}</math> | ||
Revision as of 18:57, 2 April 2022
Problem
In base 
the expanded fraction 
becomes 
, and the expanded fraction 
becomes 
. In base 
fraction , when expanded, becomes 
, while the fraction becomes 
. The sum of and , each written in the base ten, is:
Solution
First, let's write as a proper fraction in base . To do that, note that:
Multiplying this equation on both sides 
, we get:
Subtracting the first equation from the second one, we get:
Using a very similar method as above, we can see that:
and 
.
Dividing the 2 equations to get some potential solutions, we get (note that we don't have to worry about division by zero because 
):
Since non-integer bases are rarely used, we can try to assume that the solution is positive integers and see if we get any solutions.
Notice that 
has to be a negative factor of 840. We need to plug in values of 
. 840 divides -21, so we plug in 8 to check. Luckily, when 
, we see that 
, and furthermore, 11+8=19 is one of the answers. We can quickly test it into the original equations and , and we see they indeed work. Therefore the answer is 
See also
| 1966 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 38 |
Followed by Problem 40 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
