Difference between revisions of "1997 AHSME Problems/Problem 28"
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− | == See also == | + | ==Problem== |
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+ | How many ordered triples of integers <math>(a,b,c)</math> satisfy <math> |a+b|+c = 19 </math> and <math> ab+|c| = 97 </math>? | ||
+ | |||
+ | <math> \textbf{(A)}\ 0\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12 </math> | ||
+ | |||
+ | ==Solution== | ||
+ | |||
+ | [[WLOG]], let <math>a \ge 0</math>, and let <math>a \ge b</math>. We can say this because if we have one solution <math>(a,b) = (a_0, b_0)</math> with <math>a_0 \ge 0</math> and <math>a_0 > b_0</math>, we really have the four solutions <math>(a_0, b_0), (-a_0, -b_0), (b_0, a_0), (-b_0, -a_0)</math> by the symmetry of the original problem. | ||
+ | |||
+ | Furthermore, we assert that these four solutions are distinct. We can say that <math>a > b</math>, since if <math>a=b</math>, we have <math>c = 19 - 2a</math> for the first equation and either <math>c = 97 - a^2</math> or <math>c = a^2 - 97</math> for the second equation. Equating <math>19 - 2a = 97 - a^2</math> gives no integer solution, while equating <math>19 - 2a = a^2 - 97</math> also gives no integer solution. | ||
+ | |||
+ | Thus, we can now assume WLOG that <math>a \ge 0</math> and <math>a > b</math>, and each pair of <math>(a_0,b_0)</math> that we get will generate four unique solutions: <math>(a_0, b_0), (b_0, a_0), (-a_0, -b_0), (-b_0, -a_0)</math>. | ||
+ | |||
+ | |||
+ | We now divide the problem into <math>c \ge 0</math> and <math>c < 0</math>: | ||
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+ | If <math>c \ge 0</math>, we have <math>|a + b| + c = 19</math> and <math>ab + c = 97</math>. | ||
+ | |||
+ | Solving both equations for <math>c</math> and equating them, we get that <math>ab - |a + b| = 78</math>. Splitting these up, we find that either <math>ab - a - b = 78</math> or <math>ab + a+ b = 78</math>. Factoring both with [[SFFT]] gives <math>(a-1)(b-1) = 79</math> or <math>(a +1)(b+1) = 79</math>. We factor with the restrctions that <math>a \ge 0</math> and <math>a > b</math>. Since <math>79</math> is prime, we have: | ||
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+ | <math>a - 1 = 79</math> and <math>b - 1 = 1</math>, which leads to <math>(80,2)</math>. | ||
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+ | <math>a + 1 = 79</math> and <math>b + 1 = 1</math>, which leads to <math>(78, 0)</math>. | ||
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+ | Each of those solutions could generate <math>3</math> more solutions, giving a total of <math>8</math> potential solutions. However, in each the first set of four solutions, we have <math>|a + b| = 82</math>, which from the original first equation <math>|a + b| + c = 19</math> gives <math>c = 19 - 82</math>, which contradicts our initial assumption that <math>c\ge 0</math>. Similarly, for the second set of four solutions, we have <math>|a + b| = 78</math>, which leads to <math>c = 19 - 78</math>, also contradicting <math>c \ge 0</math>. | ||
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+ | |||
+ | If <math>c < 0</math>, we have <math>|a + b| + c = 19</math> and <math>ab - c = 97</math>. We note that <math>19 - c</math> must be positive whenever <math>c</math> is negative, and thus <math>|a + b| = a + b</math>. | ||
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+ | Solving both equations for <math>c</math> and using SFFT as above gives <math>(a+1)(b+1) = 117</math>. Since <math>117 = 3^2\cdot 39</math>, we factor with the restriction that <math>a\ge 0</math> and <math>a > b</math>. Thus, we can let <math>a+1 \in \{117, 39, 13\}</math>, which means <math>a \in \{116, 38, 12\}</math>. These give corresponding <math>b+1\in \{1, 3, 9\}</math>, which leads to corresponding <math>b \in \{0, 2, 8\}</math>. Combining the solutions, we have <math>(a,b) = (116, 0), (38, 2), (12, 8)</math>. | ||
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+ | Each of these three solutions permutes, negates, and permute-negates into <math>4</math> solutions as described in the start of the solution, for a total of <math>12</math> solutions. | ||
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+ | Checking our solutions to ensure <math>c < 0</math>, we find in the first set of four solutions, <math>|a + b| = 116</math>, and thus <math>c = 19 - 116</math>, which is indeed negative. | ||
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+ | In the second set of four solutions, <math>|a + b| = 40</math>, which leads to <math>c = 19 - 40</math>, which is also negative. | ||
+ | |||
+ | Finally, in the third set of four solutoins, <math>|a + b| = 20</math>, which leads to <math>c = 19 - 20</math>, which is negative. | ||
+ | |||
+ | Thus, there are <math>12</math> ordered triples, and the answer is <math>\boxed{E}</math>. | ||
+ | |||
+ | ==See also== | ||
+ | |||
{{AHSME box|year=1997|num-b=27|num-a=29}} | {{AHSME box|year=1997|num-b=27|num-a=29}} |
Revision as of 16:03, 23 August 2011
Problem
How many ordered triples of integers satisfy and ?
Solution
WLOG, let , and let . We can say this because if we have one solution with and , we really have the four solutions by the symmetry of the original problem.
Furthermore, we assert that these four solutions are distinct. We can say that , since if , we have for the first equation and either or for the second equation. Equating gives no integer solution, while equating also gives no integer solution.
Thus, we can now assume WLOG that and , and each pair of that we get will generate four unique solutions: .
We now divide the problem into and :
If , we have and .
Solving both equations for and equating them, we get that . Splitting these up, we find that either or . Factoring both with SFFT gives or . We factor with the restrctions that and . Since is prime, we have:
and , which leads to .
and , which leads to .
Each of those solutions could generate more solutions, giving a total of potential solutions. However, in each the first set of four solutions, we have , which from the original first equation gives , which contradicts our initial assumption that . Similarly, for the second set of four solutions, we have , which leads to , also contradicting .
If , we have and . We note that must be positive whenever is negative, and thus .
Solving both equations for and using SFFT as above gives . Since , we factor with the restriction that and . Thus, we can let , which means . These give corresponding , which leads to corresponding . Combining the solutions, we have .
Each of these three solutions permutes, negates, and permute-negates into solutions as described in the start of the solution, for a total of solutions.
Checking our solutions to ensure , we find in the first set of four solutions, , and thus , which is indeed negative.
In the second set of four solutions, , which leads to , which is also negative.
Finally, in the third set of four solutoins, , which leads to , which is negative.
Thus, there are ordered triples, and the answer is .
See also
1997 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 27 |
Followed by Problem 29 | |
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All AHSME Problems and Solutions |