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>. | ||
− | For <math>i^{x}=(1+i)^{y}</math>, this only occurs at <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 zero and nineteen, <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\ | + | For <math>i^{x}=(1+i)^{y}</math>, this only occurs at <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 zero and nineteen, <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\ | + | 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>(1+i)^{y}=z\neq i^x</math>, this occurs at <math>1</math> and <math>16</math>. <math>(1+i)^{y}=1</math> and <math>z=1</math> both have one solution while <math>i^{x}\neq 1</math> has fifteen solutions. <math>(1+i)^{y}=16</math> and <math>z=16</math> both have one solution, namely, <math>y=8</math> and <math>z=16</math>, while <math>i^{x}\neq 16</math> has twenty solutions. So we have <math>15\ | + | For <math>(1+i)^{y}=z\neq i^x</math>, this occurs at <math>1</math> and <math>16</math>. <math>(1+i)^{y}=1</math> and <math>z=1</math> both have one solution while <math>i^{x}\neq 1</math> has fifteen solutions. <math>(1+i)^{y}=16</math> and <math>z=16</math> both have one solution, namely, <math>y=8</math> and <math>z=16</math>, while <math>i^{x}\neq 16</math> has twenty solutions (<math>i^x</math> only cycles as <math>1, i, -1, -i</math>). So we have <math>15\cdot 1\cdot 1+20\cdot 1\cdot 1=35</math> ordered triples. |
− | In total we have <math>{95+95+35= | + | In total we have <math>{95+95+35=\boxed{\text{(D) }225}}</math> ordered triples |
== See also == | == See also == | ||
{{AMC12 box|year=2010|num-b=14|num-a=16|ab=B}} | {{AMC12 box|year=2010|num-b=14|num-a=16|ab=B}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:28, 9 October 2017
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
.
For , this only occurs at
.
has only one solution, namely,
.
has five solutions between zero and nineteen,
, 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
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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