Difference between revisions of "2010 AMC 12B Problems/Problem 15"
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We have either <math>i^{x}=(1+i)^{y}\neq z</math>, <math>i^{x}=z\neq(1+i)^{y}</math>, or <math>(1+i)^{y}=z\neq i^x</math>. | We have either <math>i^{x}=(1+i)^{y}\neq z</math>, <math>i^{x}=z\neq(1+i)^{y}</math>, or <math>(1+i)^{y}=z\neq i^x</math>. | ||
− | + | <math>i^{x}=(1+i)^{y}</math> only occurs when it is <math>1</math>. <math>(1+i)^{y}=1</math> has only one solution, namely, <math>y=0</math>. <math>i^{x}=1</math> has five solutions between 0 and 19, <math>x=0, x=4, x=8, x=12</math>, and <math>x=16</math>. <math>z\neq 1</math> has nineteen integer solutions between zero and nineteen. So for <math>i^{x}=(1+i)^{y}\neq z</math>, we have <math>5\cdot 1\cdot 19=95</math> ordered triples. | |
For <math>i^{x}=z\neq(1+i)^{y}</math>, again this only occurs at <math>1</math>. <math>(1+i)^{y}\neq 1</math> has nineteen solutions, <math>i^{x}=1</math> has five solutions, and <math>z=1</math> has one solution, so again we have <math>5\cdot 1\cdot 19=95</math> ordered triples. | For <math>i^{x}=z\neq(1+i)^{y}</math>, again this only occurs at <math>1</math>. <math>(1+i)^{y}\neq 1</math> has nineteen solutions, <math>i^{x}=1</math> has five solutions, and <math>z=1</math> has one solution, so again we have <math>5\cdot 1\cdot 19=95</math> ordered triples. |
Latest revision as of 18:39, 23 December 2020
Problem 15
For how many ordered triples of nonnegative integers less than are there exactly two distinct elements in the set , where ?
Solution
We have either , , or .
only occurs when it is . has only one solution, namely, . has five solutions between 0 and 19, , and . has nineteen integer solutions between zero and nineteen. So for , we have ordered triples.
For , again this only occurs at . has nineteen solutions, has five solutions, and has one solution, so again we have ordered triples.
For , this occurs at and . and both have one solution while has fifteen solutions. and both have one solution, namely, and , while has twenty solutions ( only cycles as ). So we have ordered triples.
In total we have ordered triples
Small Clarification
To more clearly see why the reasoning above is true, try converting the complex numbers into exponential form. That way, we can more easily raise the numbers to , and respectively.
See also
2010 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
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 | |
All AMC 12 Problems and Solutions |
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