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Difference between revisions of "2001 AMC 8 Problems/Problem 2"

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==Problem==
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I'm thinking of two whole numbers. Their product is 24 and their sum is 11. What is the larger number?
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<math>\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 12</math>
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==Solution 1==
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Let the numbers be <math> x </math> and <math> y </math>. Then we have <math> x+y=11 </math> and <math> xy=24 </math>. Solving for <math> x </math> in the first equation yields <math> x=11-y </math>, and substituting this into the second equation gives <math> (11-y)(y)=24 </math>. Simplifying this gives <math> -y^2+11y=24 </math>, or <math> y^2-11y+24=0 </math>. This factors as <math> (y-3)(y-8)=0 </math>, so <math> y=3 </math> or <math> y=8 </math>, and the corresponding <math> x </math> values are <math> x=8 </math> and <math> x=3 </math>. These are essentially the same answer: one number is <math> 3 </math> and one number is <math> 8 </math>, so the largest number is <math> 8, \boxed{\text{D}} </math>.
 
Let the numbers be <math> x </math> and <math> y </math>. Then we have <math> x+y=11 </math> and <math> xy=24 </math>. Solving for <math> x </math> in the first equation yields <math> x=11-y </math>, and substituting this into the second equation gives <math> (11-y)(y)=24 </math>. Simplifying this gives <math> -y^2+11y=24 </math>, or <math> y^2-11y+24=0 </math>. This factors as <math> (y-3)(y-8)=0 </math>, so <math> y=3 </math> or <math> y=8 </math>, and the corresponding <math> x </math> values are <math> x=8 </math> and <math> x=3 </math>. These are essentially the same answer: one number is <math> 3 </math> and one number is <math> 8 </math>, so the largest number is <math> 8, \boxed{\text{D}} </math>.
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==Solution 2==
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Use the answers to attempt to "reverse engineer" an appropriate pair of numbers.  Looking at option <math>A</math>, guess that one of the numbers is <math>3</math>.  If the sum of two numbers is <math>11</math> and one is <math>3</math>, then other must be <math>11 - 3 = 8</math>.  The product of those numbers is <math>3\cdot 8 = 24</math>, which is the second condition of the problem, so our number are <math>3</math> and <math>8</math>.
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However, <math>3</math> is the smaller of the two numbers, so the answer is <math> 8</math> or <math>\boxed{D}</math>.
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===Solution 3===
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We go through the divisor pairs of <math>24</math> to see which two numbers sum to <math>11</math>. The numbers clearly cannot be negative. If one was negative, then the other must also be negative in order to multiply to a positive product, but it would be impossible for the numbers to add up to a positive sum. So, we look at the positive divisor pairs of <math>24</math>, namely <math>1</math> and <math>24</math>, <math>2</math> and <math>12</math>, <math>3</math> and <math>8</math>, and <math>4</math> and <math>6</math>. The only pair of numbers that sums to <math>11</math> is <math>3</math> and <math>8</math>. The larger number is <math>8</math>, so the answer is <math>\boxed{D}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2001|num-b=1|num-a=3}}
 
{{AMC8 box|year=2001|num-b=1|num-a=3}}
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{{MAA Notice}}

Latest revision as of 04:58, 8 July 2024

Problem

I'm thinking of two whole numbers. Their product is 24 and their sum is 11. What is the larger number?

$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 12$

Solution 1

Let the numbers be $x$ and $y$. Then we have $x+y=11$ and $xy=24$. Solving for $x$ in the first equation yields $x=11-y$, and substituting this into the second equation gives $(11-y)(y)=24$. Simplifying this gives $-y^2+11y=24$, or $y^2-11y+24=0$. This factors as $(y-3)(y-8)=0$, so $y=3$ or $y=8$, and the corresponding $x$ values are $x=8$ and $x=3$. These are essentially the same answer: one number is $3$ and one number is $8$, so the largest number is $8, \boxed{\text{D}}$.

Solution 2

Use the answers to attempt to "reverse engineer" an appropriate pair of numbers. Looking at option $A$, guess that one of the numbers is $3$. If the sum of two numbers is $11$ and one is $3$, then other must be $11 - 3 = 8$. The product of those numbers is $3\cdot 8 = 24$, which is the second condition of the problem, so our number are $3$ and $8$.

However, $3$ is the smaller of the two numbers, so the answer is $8$ or $\boxed{D}$.

Solution 3

We go through the divisor pairs of $24$ to see which two numbers sum to $11$. The numbers clearly cannot be negative. If one was negative, then the other must also be negative in order to multiply to a positive product, but it would be impossible for the numbers to add up to a positive sum. So, we look at the positive divisor pairs of $24$, namely $1$ and $24$, $2$ and $12$, $3$ and $8$, and $4$ and $6$. The only pair of numbers that sums to $11$ is $3$ and $8$. The larger number is $8$, so the answer is $\boxed{D}$.

See Also

2001 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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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