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

Latest revision as of 21:20, 17 October 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 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 $\boxed{\mathrm{(C)} 79}$ .

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

@@ Line 17: / Line 17: @@
-==Solution 3 making equations ==
+==Solution 3 ==
-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</math>. Now we now that <math>a^2-b^2=2a+1</math> and since <math>2a+1</math> will always be an odd number, we can narrow the answers down to (C) and (E). Since <math>2a+1=131</math> would be too big for our range of <math>a+b<100</math>, we would choose <math>\boxed{\mathrm{(C)} 79}</math>.
+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
 ==Video Solution==

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