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

Latest revision as of 01:12, 16 November 2023

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 are $a^2-b^2=(a+b)(a-b)$ , which equals to $a+b$ , because $a$ and $b$ are consecutive. And 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 that the sum of $a$ and $b$ is less than 100, you can eliminate all answers expect for $\boxed{\mathrm{(C)} 79}$ .

Video Solution

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

Video Solution by WhyMath

https://youtu.be/BrEqmDq82rw

~savannahsolver

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

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

@@ Line 1: / Line 1: @@
-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>.
+== Problem ==
-Since this is an "or" condition, that means you add the possibilities because you can win either way (same color or same matching alphabet)...
+Pick two consecutive positive integers whose sum is less than <math>100</math>. Square both
+of those integers and then find the difference of the squares. Which of the
+following could be the difference?
-<math> \frac{3}{7} + \frac{1}{7} </math> = <math> \frac{4}{7} </math> <math> \Longrightarrow </math> <math> \boxed{D} </math>
+<math>\mathrm{(A)}\ 2 \qquad \mathrm{(B)}\ 64 \qquad \mathrm{(C)}\ 79 \qquad \mathrm{(D)}\ 96 \qquad \mathrm{(E)}\ 131</math>
+== Solution 1 ==
+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.
+<math> 2x+1=131 </math> contradicts the fact that <math> 2x+1<100 </math>, so the answer is <math> \boxed{\mathrm{(C)} 79} </math>
+==Solution 2==
+Since for two consecutive numbers <math>a</math> and <math>b</math>, the difference between their squares are <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. And 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 that the sum of <math>a</math> and <math>b</math> is less than 100, you can eliminate all answers expect for <math>\boxed{\mathrm{(C)} 79}</math>.
+==Video Solution==
+https://youtu.be/8FGl2vtO1vs Soo, DRMS, NM
+==Video Solution by WhyMath==
+https://youtu.be/BrEqmDq82rw
+~savannahsolver
+==See Also==
+{{AMC8 box|year=2007|num-b=18|num-a=20}}
+{{MAA Notice}}

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