Difference between revisions of "2006 SMT/Algebra Problems/Problem 6"
(Created page with "==Problem== Let <math> a, b, c </math> be real numbers satisfying: <cmath> ab-a=b+119 </cmath> <cmath> bc-b=c+59 </cmath> <cmath> ca-c=a+71 </cmath> Determine all possible valu...") |
(Added second solution.) |
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Therefore, the possible values of <math> a+b+c </math> are <math> \boxed{31 \text{ and } -25} </math>. | Therefore, the possible values of <math> a+b+c </math> are <math> \boxed{31 \text{ and } -25} </math>. | ||
+ | ==Solution 2== | ||
+ | We can rearrange the equations as follows: | ||
+ | |||
+ | <math>ab-a-b = 119</math> | ||
+ | |||
+ | <math>bc-b-c = 71</math> | ||
+ | |||
+ | <math>ca-c-a = 59</math> | ||
+ | |||
+ | Then, using [[Simon's Favorite Factoring Trick]] we get: | ||
+ | |||
+ | <math>(a-1)(b-1) = 120</math> | ||
+ | |||
+ | <math>(b-1)(c-1)= 72</math> | ||
+ | |||
+ | <math>(c-1)(a-1) = 60</math> | ||
+ | |||
+ | Multiplying the three equations together yields <math>[(a-1)(b-1)(c-1)]^2 = 120 \cdot 72 \cdot 60 = 518400 \leadsto (a-1)(b-1)(c-1) = \pm 720</math> | ||
+ | |||
+ | If <math>(a-1)(b-1)(c-1) = +720</math>, then dividing this equation by the factored equations yields: | ||
+ | |||
+ | <math>a-1 = \tfrac{720}{72} = 10 \leadsto a = 11</math> | ||
+ | |||
+ | <math>b-1 = \tfrac{720}{60} = 12 \leadsto b = 13</math> | ||
+ | |||
+ | <math>c-1 = \tfrac{720}{120} = 6 \leadsto c = 7</math> | ||
+ | |||
+ | and <math>a+b+c = 11+13+7 = 31</math> | ||
+ | |||
+ | If <math>(a-1)(b-1)(c-1) = -720</math>, then dividing this equation by the factored equations yields: | ||
+ | |||
+ | <math>a-1 = \tfrac{-720}{72} = -10 \leadsto a = -9</math> | ||
+ | |||
+ | <math>b-1 = \tfrac{-720}{60} = -12 \leadsto b = -11</math> | ||
+ | |||
+ | <math>c-1 = \tfrac{-720}{120} = -6 \leadsto c = -5</math> | ||
+ | |||
+ | and <math>a+b+c = -9-11-5 = -25</math> | ||
+ | |||
+ | Thus, <math>a+b+c = \boxed{31 \text{ \ or \ } -25}</math>. | ||
==See Also== | ==See Also== | ||
[[2006 SMT/Algebra Problems]] | [[2006 SMT/Algebra Problems]] |
Latest revision as of 00:24, 31 May 2012
Contents
[hide]Problem
Let be real numbers satisfying:
Determine all possible values of .
Solution
From the first equation, we have . Plugging this into the third equation, we get . Multiplying both sides by , we get .
Now we plug that into the second equation. We have . Getting rid of the fractions, we have . We can factor that as , so or .
If , then and , so .
If , then and , so .
Therefore, the possible values of are .
Solution 2
We can rearrange the equations as follows:
Then, using Simon's Favorite Factoring Trick we get:
Multiplying the three equations together yields
If , then dividing this equation by the factored equations yields:
and
If , then dividing this equation by the factored equations yields:
and
Thus, .