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Difference between revisions of "2007 AMC 8 Problems/Problem 19"

(Created page with 'Suppose you take out a card: Green, <math> A </math>. There are <math> 7 </math> cards left in the deck. There are three cards with the same color as the first card: Green <math>…')
 
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Suppose you take out a card: Green, <math> A </math>. There are <math> 7 </math> cards left in the deck. There are three cards with the same color as the first card: Green <math> B </math>, Green <math> C </math>, Green <math> D </math>. There is only <math> 1 </math> card with the matching alphabet: Red <math> A </math>.
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== Problem ==
  
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Since this is an "or" condition, that means you add the possibilities because you can win either way (same color or same matching alphabet)...
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Pick two consecutive positive integers whose sum is less than <math>100</math>. Square both
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of those integers and then find the difference of the squares. Which of the
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following could be the difference?
  
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<math> \frac{3}{7} + \frac{1}{7} </math> = <math> \frac{4}{7} </math> <math> \Longrightarrow </math> <math> \boxed{D} </math>
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<math>\mathrm{(A)}\ 2 \qquad \mathrm{(B)}\ 64 \qquad \mathrm{(C)}\ 79 \qquad \mathrm{(D)}\ 96 \qquad \mathrm{(E)}\ 131</math>
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== Solution ==
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Let the smaller of the two numbers be <math>x</math>. Then, the problem states that <math> 2x+1<100</math>. <math> (x+1)^2-x^2=x^2+2x+1-x^2=2x+1 </math>. <math> 2x+1 </math> is obviously odd, so only answer choices C and E need to be considered.
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<math> 2x+1=131 </math> refutes the fact that <math> 2x+1<100 </math>, so the answer is <math> \mathrm{(C)}\ 79 </math>

Revision as of 22:57, 14 November 2011

Problem

Pick two consecutive positive integers whose sum is less than $100$. Square both of those integers and then find the difference of the squares. Which of the following could be the difference?

$\mathrm{(A)}\ 2 \qquad \mathrm{(B)}\ 64 \qquad \mathrm{(C)}\ 79 \qquad \mathrm{(D)}\ 96 \qquad \mathrm{(E)}\ 131$

Solution

Let the smaller of the two numbers be $x$. Then, the problem states that $2x+1<100$. $(x+1)^2-x^2=x^2+2x+1-x^2=2x+1$. $2x+1$ is obviously odd, so only answer choices C and E need to be considered.

$2x+1=131$ refutes the fact that $2x+1<100$, so the answer is $\mathrm{(C)}\ 79$

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