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

Revision as of 04:50, 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 (quick number theory)

We can go through the divisor pairs of $24$ and see which two numbers sum to $11$ . The numbers clearly cannot be negative. If one were negative, then both must be negative to multiply to a positive product, and it would be impossible for the numbers to add up to a positive sum. So, we look at the positive divisor pairs of $24$ . These pairs are $1$ and $24$ , $2$ and $12$ , $3$ and $8$ , and $4$ and $6$ . Only one of these adds to $11$ , that being $3$ and $8$ . The larger number of the pair is $8$ , or $\boxed{D}$ .

@@ Line 17: / Line 17: @@
 ===Solution 3 (quick number theory)===
-We can go through the divisor pairs of <math>24</math> and see which two numbers sum to <math>11</math>. The numbers clearly cannot be negative. If one were negative, then both must be negative to multiply to a positive product, and 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>. These pairs are <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>. Only one of these adds to <math>11</math> - <math>3</math> and <math>8</math>. The larger number of the pair is <math>8</math>, or <math>\boxed{D}</math>.
+We can go through the divisor pairs of <math>24</math> and see which two numbers sum to <math>11</math>. The numbers clearly cannot be negative. If one were negative, then both must be negative to multiply to a positive product, and 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>. These pairs are <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>. Only one of these adds to <math>11</math>, that being <math>3</math> and <math>8</math>. The larger number of the pair is <math>8</math>, or <math>\boxed{D}</math>.
 ==See Also==
 {{AMC8 box|year=2001|num-b=1|num-a=3}}
 {{MAA Notice}}

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