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

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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 1 ==
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Let the smaller of the two numbers be <math>x</math>. Then, the problem states that <math> (x+1)+x<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> contradicts the fact that <math> 2x+1<100 </math>, so the answer is <math> \boxed{\mathrm{(C)} 79} </math>
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==Solution 2==
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Since for two consecutive numbers <math>a</math> and <math>b</math>, the difference between their squares is <math>a^2-b^2=(a+b)(a-b)</math>, which equals to <math>a+b</math>, because <math>a</math> and <math>b</math> are consecutive. Because they are consecutive, one number must be even, and the other odd. Since the sum of an even and an odd number is always odd, and the sum of <math>a</math> and <math>b</math> is less than 100, you can eliminate all answers except for <math>\boxed{\mathrm{(C)} 79}</math>.
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==Solution 3 ==
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From the question we can make <math>a^2-b^2</math> and we can let <math>a</math> be <math>b-1</math> where <math>a+b<100</math>. First we factor <math>a^2-b^2</math> to <math>(a+b)(a-b)</math> and then plug in <math>a=b-1</math> to have: <math>(a+a+1)(a-a+1)=(2a+1)(1)=2a+1=a+b</math>. since <math>a+b</math> will always be an odd number, we can narrow the answers down to (C) and (E). Since <math>131</math> would be too big for our range of <math>a+b<100</math>, we would choose <math>\boxed{\mathrm{(C)} 79}</math>. note that this solution would be better if the question was asking for the values of a and b
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==Video Solution==
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https://youtu.be/8FGl2vtO1vs Soo, DRMS, NM
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==Video Solution by WhyMath==
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https://youtu.be/BrEqmDq82rw
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~savannahsolver
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==See Also==
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{{AMC8 box|year=2007|num-b=18|num-a=20}}
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{{MAA Notice}}

Latest revision as of 11:55, 24 December 2024

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 1

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

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

Solution 2

Since for two consecutive numbers $a$ and $b$, the difference between their squares is $a^2-b^2=(a+b)(a-b)$, which equals to $a+b$, because $a$ and $b$ are consecutive. Because they are consecutive, one number must be even, and the other odd. Since the sum of an even and an odd number is always odd, and the sum of $a$ and $b$ is less than 100, you can eliminate all answers except for $\boxed{\mathrm{(C)} 79}$.

Solution 3

From the question we can make $a^2-b^2$ and we can let $a$ be $b-1$ where $a+b<100$. First we factor $a^2-b^2$ to $(a+b)(a-b)$ and then plug in $a=b-1$ to have: $(a+a+1)(a-a+1)=(2a+1)(1)=2a+1=a+b$. since $a+b$ will always be an odd number, we can narrow the answers down to (C) and (E). Since $131$ would be too big for our range of $a+b<100$, we would choose $\boxed{\mathrm{(C)} 79}$. note that this solution would be better if the question was asking for the values of a and b

Video Solution

https://youtu.be/8FGl2vtO1vs Soo, DRMS, NM

Video Solution by WhyMath

https://youtu.be/BrEqmDq82rw

~savannahsolver

See Also

2007 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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 AJHSME/AMC 8 Problems and Solutions

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