Difference between revisions of "2021 Fall AMC 12B Problems/Problem 7"

Revision as of 16:16, 24 November 2021

The following problem is from both the 2021 Fall AMC 10B #12 and 2021 Fall AMC 12B #7, so both problems redirect to this page.

Problem

Which of the following conditions is sufficient to guarantee that integers $x$ , $y$ , and $z$ satisfy the equation $x(x-y)+y(y-z)+z(z-x) = 1?$

$\textbf{(A)} \: x>y$ and $y=z$

$\textbf{(B)} \: x=y-1$ and $y=z-1$

$\textbf{(C)} \: x=z+1$ and $y=x+1$

$\textbf{(D)} \: x=z$ and $y-1=x$

$\textbf{(E)} \: x+y+z=1$

Solution 1 (Strategy)

Looking at the answer choices and the question, the simplest ones to plug in would be equalities because it would make one term of the equation become zero. We see that answer choices A and D have the simplest equalities in them. However, A has an inequality too, so it would be simpler to plug in D which has another equality. We see that $x=z$ and $y-1=x$ means the equation becomes $x(x-(x+1)) + (x+1)(x+1 - x) = 1 \implies -x + x + 1 = 1 \implies 1 = 1$ , which is always true, so the answer is $\boxed{D}$

Solution 2

Plugging in every choice, we see that choice $\textbf{(D)}$ works.

We have $y=x+1, z=x$ , so $x(x-y)+y(y-z)+z(z-x)=x(x-(x+1))+(x+1)((x+1)-x)+x(x-x)=x(-1)+(x+1)(1)=1.$ Our answer is $\textbf{(D)}$ .

~kingofpineapplz

Solution 3 (Bash)

Just plug in all these options one by one, and one sees that all but $D$ fails to satisfy the equation.

For $D$ , substitute $z=x$ and $y=x+1$ :

$LHS=x(x-(x+1))+(x+1)(x+1-x)+x(x-x)=(-x)+(x+1)=1=RHS$

Hence the answer is $\boxed{\textbf{(D)}}.$

~Wilhelm Z

2021 Fall AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 16: / Line 16: @@
 ==Solution 1 (Strategy)==
-Looking at the answer choices and the question, the simplest ones to plug in would be equalities because it would make one term of the equation become zero. We see that answer choices A and D have the simplest equalities in them. However, A has an inequality too, so it would be simpler to plug in D which has another equality. We see that <math>x=z</math> and <math>y-1=x</math> satisfy the equation, so the answer is <math>\boxed{D}</math>
+Looking at the answer choices and the question, the simplest ones to plug in would be equalities because it would make one term of the equation become zero. We see that answer choices A and D have the simplest equalities in them. However, A has an inequality too, so it would be simpler to plug in D which has another equality. We see that <math>x=z</math> and <math>y-1=x</math> means the equation becomes <math>x(x-(x+1)) + (x+1)(x+1 - x)  = 1 \implies -x + x + 1 = 1 \implies 1 =  1</math>, which is always true, so the answer is <math>\boxed{D}</math>
 ==Solution 2==
 Plugging in every choice, we see that choice <math>\textbf{(D)}</math> works.

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