Difference between revisions of "1966 AHSME Problems/Problem 39"
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== Solution == | == Solution == | ||
| − | <math>\ | + | First, let's write <math>F_1</math> as a proper fraction in base <math>R_1</math>. To do that, note that: |
| + | <math>F_1=0.373737\dots</math> | ||
| + | Multiplying this equation on both sides <math>R_1^2</math>, we get: | ||
| + | <math>R_1^2F_1=37.373737\dots</math> | ||
| + | Subtracting the first equation from the second one, we get: | ||
| + | <math>R_1^2F_1-F_1=37\\F_1(R_1^2-1)=37\\F_1=\frac{3R_1+7}{R_1^2-1}</math> | ||
| + | Using a very similar method as above, we can see that: | ||
| + | <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>): | ||
| + | <math>\frac{3R_1+7}{7R_1+3}=\frac{2R_2+5}{5R_2+2}\\15R_1R_2+6R_1+35R_2+14=14R_1R_2+35R_1+6R_2+15\\R_1R_2-29R_1+29R_2=1\\\left(R_1+29\right)\left(R_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. | ||
| + | Notice that <math>(R_2-29)</math> has to be a negative factor of 840. We need to plug in values of <math>R_2 > 7</math>. 840 divides -21, so we plug in 8 to check. Luckily, when <math>R_2 = 8</math>, we see that <math>R_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)19}</math> | ||
| + | |||
| + | ~ Minor edits by Raghav(AOPS - CatalanThinker) | ||
== See also == | == See also == | ||
| − | {{AHSME box|year=1966|num-b=38|num-a=40}} | + | {{AHSME 40p box|year=1966|num-b=38|num-a=40}} |
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 06:57, 4 August 2025
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 
~ Minor edits by Raghav(AOPS - CatalanThinker)
See also
| 1966 AHSC (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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
