Difference between revisions of "1987 AHSME Problems/Problem 10"
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== Solution == | == Solution == | ||
− | 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 = | + | 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 of non-zero real numbers have the property that each number is the product of the other two?
Solution
We have , , and , so multiplying these three equations together gives , and as , , and are all non-zero, we cannot have , so we must have . Now substituting gives . If , then the system becomes , so either or , giving solutions. If , the system becomes , so or , giving another solutions. Thus the total number of solutions is , which is answer .
See also
1987 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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All AHSME Problems and Solutions |
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