Difference between revisions of "2012 AMC 10A Problems/Problem 24"

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== Problem 24 ==
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{{duplicate|[[2012 AMC 12A Problems|2012 AMC 12A #21]] and [[2012 AMC 10A Problems|2012 AMC 10A #24]]}}
  
Let <math>a</math>, <math>b</math>, and <math>c</math> be positive integers with <math>a\ge</math> <math>b\ge</math> <math>c</math> such that
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== Problem ==
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<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Let <math>a</math>, <math>b</math>, and <math>c</math> be positive integers with <math>a\ge</math> <math>b\ge</math> <math>c</math> such that
 
<math>a^2-b^2-c^2+ab=2011</math> and
 
<math>a^2-b^2-c^2+ab=2011</math> and
 
<math>a^2+3b^2+3c^2-3ab-2ac-2bc=-1997</math>.
 
<math>a^2+3b^2+3c^2-3ab-2ac-2bc=-1997</math>.
  
What is <math>a</math>?
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What is <math>a</math>?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude>
  
 
<math> \textbf{(A)}\ 249\qquad\textbf{(B)}\ 250\qquad\textbf{(C)}\ 251\qquad\textbf{(D)}\ 252\qquad\textbf{(E)}\ 253 </math>
 
<math> \textbf{(A)}\ 249\qquad\textbf{(B)}\ 250\qquad\textbf{(C)}\ 251\qquad\textbf{(D)}\ 252\qquad\textbf{(E)}\ 253 </math>
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== Solution ==
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Add the two equations.
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<math>2a^2 + 2b^2 + 2c^2 - 2ab - 2ac - 2bc = 14</math>.
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Now, this can be rearranged and factored.
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<math>(a^2 - 2ab + b^2) + (a^2 - 2ac + c^2) + (b^2 - 2bc + c^2) = 14</math>
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<math>(a - b)^2 + (a - c)^2 + (b - c)^2 = 14</math>
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<math>a</math>, <math>b</math>, and <math>c</math> are all integers, so the three terms on the left side of the equation must all be perfect squares.
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We see that the only is possibility is <math>14 = 9 + 4 + 1</math>.
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<math>(a-c)^2 = 9 \Rightarrow a-c = 3</math>, since <math>a-c</math> is the biggest difference. It is impossible to determine by inspection whether <math>a-b = 1</math> or <math>2</math>, or whether <math>b-c = 1</math> or <math>2</math>.
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We want to solve for <math>a</math>, so take the two cases and solve them each for an expression in terms of <math>a</math>. Our two cases are <math>(a, b, c) = (a, a-1, a-3)</math> or <math>(a, a-2, a-3)</math>. Plug these values into one of the original equations to see if we can get an integer for <math>a</math>.
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<math>a^2 - (a-1)^2 - (a-3)^2 + a(a-1) = 2011</math>, after some algebra, simplifies to
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<math>7a = 2021</math>. <math>2021</math> is not divisible by <math>7</math>, so <math>a</math> is not an integer.
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The other case gives <math>a^2 - (a-2)^2 - (a-3)^2 + a(a-2) = 2011</math>, which simplifies to <math>8a = 2024</math>. Thus, <math>a = 253</math> and the answer is <math>\boxed{\textbf{(E)}\ 253}</math>.
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==Video Solution by Richard Rusczyk==
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https://artofproblemsolving.com/videos/amc/2012amc12a/250
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~dolphin7
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== See Also ==
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{{AMC10 box|year=2012|ab=A|num-b=23|num-a=25}}
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{{AMC12 box|year=2012|ab=A|num-b=20|num-a=22}}
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[[Category:Introductory Algebra Problems]]
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[[Category:Algebraic Manipulations Problems]]
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{{MAA Notice}}

Latest revision as of 16:55, 20 January 2021

The following problem is from both the 2012 AMC 12A #21 and 2012 AMC 10A #24, so both problems redirect to this page.

Problem

Let $a$, $b$, and $c$ be positive integers with $a\ge$ $b\ge$ $c$ such that $a^2-b^2-c^2+ab=2011$ and $a^2+3b^2+3c^2-3ab-2ac-2bc=-1997$.

What is $a$?

$\textbf{(A)}\ 249\qquad\textbf{(B)}\ 250\qquad\textbf{(C)}\ 251\qquad\textbf{(D)}\ 252\qquad\textbf{(E)}\ 253$

Solution

Add the two equations.

$2a^2 + 2b^2 + 2c^2 - 2ab - 2ac - 2bc = 14$.

Now, this can be rearranged and factored.

$(a^2 - 2ab + b^2) + (a^2 - 2ac + c^2) + (b^2 - 2bc + c^2) = 14$

$(a - b)^2 + (a - c)^2 + (b - c)^2 = 14$

$a$, $b$, and $c$ are all integers, so the three terms on the left side of the equation must all be perfect squares. We see that the only is possibility is $14 = 9 + 4 + 1$.

$(a-c)^2 = 9 \Rightarrow a-c = 3$, since $a-c$ is the biggest difference. It is impossible to determine by inspection whether $a-b = 1$ or $2$, or whether $b-c = 1$ or $2$.

We want to solve for $a$, so take the two cases and solve them each for an expression in terms of $a$. Our two cases are $(a, b, c) = (a, a-1, a-3)$ or $(a, a-2, a-3)$. Plug these values into one of the original equations to see if we can get an integer for $a$.

$a^2 - (a-1)^2 - (a-3)^2 + a(a-1) = 2011$, after some algebra, simplifies to $7a = 2021$. $2021$ is not divisible by $7$, so $a$ is not an integer.

The other case gives $a^2 - (a-2)^2 - (a-3)^2 + a(a-2) = 2011$, which simplifies to $8a = 2024$. Thus, $a = 253$ and the answer is $\boxed{\textbf{(E)}\ 253}$.

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2012amc12a/250

~dolphin7

See Also

2012 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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
2012 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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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