Difference between revisions of "1997 AHSME Problems/Problem 28"

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== See also ==
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==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>?
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<math> \textbf{(A)}\ 0\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12 </math>
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==Solution==
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[[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. 
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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.
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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>.
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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>.
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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.
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Finally, in the third set of four solutoins, <math>|a + b| = 20</math>, which leads to <math>c = 19 - 20</math>, which is negative.
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Thus, there are <math>12</math> ordered triples, and the answer is <math>\boxed{E}</math>.
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==See also==
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{{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 $(a,b,c)$ satisfy $|a+b|+c = 19$ and $ab+|c| = 97$?

$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12$

Solution

WLOG, let $a \ge 0$, and let $a \ge b$. We can say this because if we have one solution $(a,b) = (a_0, b_0)$ with $a_0 \ge 0$ and $a_0 > b_0$, we really have the four solutions $(a_0, b_0), (-a_0, -b_0), (b_0, a_0), (-b_0, -a_0)$ by the symmetry of the original problem.

Furthermore, we assert that these four solutions are distinct. We can say that $a > b$, since if $a=b$, we have $c = 19 - 2a$ for the first equation and either $c = 97 - a^2$ or $c = a^2 - 97$ for the second equation. Equating $19 - 2a = 97 - a^2$ gives no integer solution, while equating $19 - 2a = a^2 - 97$ also gives no integer solution.

Thus, we can now assume WLOG that $a \ge 0$ and $a > b$, and each pair of $(a_0,b_0)$ that we get will generate four unique solutions: $(a_0, b_0), (b_0, a_0), (-a_0, -b_0), (-b_0, -a_0)$.


We now divide the problem into $c \ge 0$ and $c < 0$:

If $c \ge 0$, we have $|a + b| + c = 19$ and $ab + c = 97$.

Solving both equations for $c$ and equating them, we get that $ab - |a + b| = 78$. Splitting these up, we find that either $ab - a - b = 78$ or $ab + a+ b = 78$. Factoring both with SFFT gives $(a-1)(b-1) = 79$ or $(a +1)(b+1) = 79$. We factor with the restrctions that $a \ge 0$ and $a > b$. Since $79$ is prime, we have:

$a - 1 = 79$ and $b - 1 = 1$, which leads to $(80,2)$.

$a + 1 = 79$ and $b + 1 = 1$, which leads to $(78, 0)$.

Each of those solutions could generate $3$ more solutions, giving a total of $8$ potential solutions. However, in each the first set of four solutions, we have $|a + b| = 82$, which from the original first equation $|a + b| + c = 19$ gives $c = 19 - 82$, which contradicts our initial assumption that $c\ge 0$. Similarly, for the second set of four solutions, we have $|a + b| = 78$, which leads to $c = 19 - 78$, also contradicting $c \ge 0$.


If $c < 0$, we have $|a + b| + c = 19$ and $ab - c = 97$. We note that $19 - c$ must be positive whenever $c$ is negative, and thus $|a + b| = a + b$.

Solving both equations for $c$ and using SFFT as above gives $(a+1)(b+1) = 117$. Since $117 = 3^2\cdot 39$, we factor with the restriction that $a\ge 0$ and $a > b$. Thus, we can let $a+1  \in \{117, 39, 13\}$, which means $a \in \{116, 38, 12\}$. These give corresponding $b+1\in \{1, 3, 9\}$, which leads to corresponding $b \in \{0, 2, 8\}$. Combining the solutions, we have $(a,b) = (116, 0), (38, 2), (12, 8)$.

Each of these three solutions permutes, negates, and permute-negates into $4$ solutions as described in the start of the solution, for a total of $12$ solutions.

Checking our solutions to ensure $c < 0$, we find in the first set of four solutions, $|a + b| = 116$, and thus $c = 19 - 116$, which is indeed negative.

In the second set of four solutions, $|a + b| = 40$, which leads to $c = 19 - 40$, which is also negative.

Finally, in the third set of four solutoins, $|a + b| = 20$, which leads to $c = 19 - 20$, which is negative.

Thus, there are $12$ ordered triples, and the answer is $\boxed{E}$.

See also

1997 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 27
Followed by
Problem 29
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