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Difference between revisions of "1987 AHSME Problems/Problem 10"

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== Solution ==
 
== Solution ==
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We have <math>ab = c</math>, <math>bc = a</math>, and <math>ca = b</math>, so multiplying these three equations together gives <math>a^{2}b^{2}c^{2} = abc \implies abc(abc-1)=0</math>, and as <math>a</math>, <math>b</math>, and <math>c</math> are all non-zero, we cannot have <math>abc = 0</math>, so we must have <math>abc = 1</math>. Now substituting <math>bc = a</math> gives <math>a(bc) = 1 \implies a^2 = 1 \implies a = \pm 1</math>. If <math>a = 1</math>, then the system becomes <math>b = c, bc = 1, c = b</math>, so either <math>b = c = 1</math> or <math>b = c = -1</math>, giving <math>2</math> solutions. If <math>a = -1</math>, the system becomes <math>-b = c, bc = -1, -c = b</math>, so again <math>b = c = 1</math> or <math>b = c = -1</math>, giving another <math>2</math> solutions. Thus the total number of solutions is <math>2 + 2 = 4</math>, which is answer <math>\boxed{\text{D}}</math>.
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We have <math>ab = c</math>, <math>bc = a</math>, and <math>ca = b</math>, so multiplying these three equations together gives <math>a^{2}b^{2}c^{2} = abc \implies abc(abc-1)=0</math>, and as <math>a</math>, <math>b</math>, and <math>c</math> are all non-zero, we cannot have <math>abc = 0</math>, so we must have <math>abc = 1</math>. Now substituting <math>bc = a</math> gives <math>a(bc) = 1 \implies a^2 = 1 \implies a = \pm 1</math>. If <math>a = 1</math>, then the system becomes <math>b = c, bc = 1, c = b</math>, so either <math>b = c = 1</math> or <math>b = c = -1</math>, giving <math>2</math> solutions. If <math>a = -1</math>, the system becomes <math>-b = c, bc = -1, -c = b</math>, so <math>-b = c = 1</math> or <math>b = -c = 1</math>, giving another <math>2</math> solutions. Thus the total number of solutions is <math>2 + 2 = 4</math>, which is answer <math>\boxed{\text{D}}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 17:52, 20 August 2020

Problem

How many ordered triples $(a, b, c)$ of non-zero real numbers have the property that each number is the product of the other two?

$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$

Solution

We have $ab = c$, $bc = a$, and $ca = b$, so multiplying these three equations together gives $a^{2}b^{2}c^{2} = abc \implies abc(abc-1)=0$, and as $a$, $b$, and $c$ are all non-zero, we cannot have $abc = 0$, so we must have $abc = 1$. Now substituting $bc = a$ gives $a(bc) = 1 \implies a^2 = 1 \implies a = \pm 1$. If $a = 1$, then the system becomes $b = c, bc = 1, c = b$, so either $b = c = 1$ or $b = c = -1$, giving $2$ solutions. If $a = -1$, the system becomes $-b = c, bc = -1, -c = b$, so $-b = c = 1$ or $b = -c = 1$, giving another $2$ solutions. Thus the total number of solutions is $2 + 2 = 4$, which is answer $\boxed{\text{D}}$.

See also

1987 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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

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