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

Revision as of 17:06, 27 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

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 2 (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

Solution 3 (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 $\boxed{D}$

~KingRavi

Solution 4

By trying conditions in each choice, we find the correct answer is $\boxed{\textbf{(D) }x = z \mbox{ and } y - 1 = x}$ .

~Steven Chen (www.professorchenedu.com)

2021 Fall AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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 10 Problems and Solutions

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 47: / Line 47: @@
 ~Steven Chen (www.professorchenedu.com)
+{{AMC10 box|year=2021 Fall|ab=B|num-a=13|num-b=11}}
 {{AMC12 box|year=2021 Fall|ab=B|num-a=8|num-b=6}}
 {{MAA Notice}}

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