Difference between revisions of "2001 AMC 8 Problems/Problem 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>. | 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>. | ||
− | However, <math>3</math> is the smaller of the two numbers, so the answer is <math> 8</math> or <math>\boxed{D}</math>. | + | However, <math>3</math> is the smaller of the two numbers, so the answer is <math> 8</math> or <math>\boxed{D}</math>. |
+ | |||
+ | ===Solution 3=== | ||
+ | |||
+ | 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}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 05:58, 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?
Solution 1
Let the numbers be and
. Then we have
and
. Solving for
in the first equation yields
, and substituting this into the second equation gives
. Simplifying this gives
, or
. This factors as
, so
or
, and the corresponding
values are
and
. These are essentially the same answer: one number is
and one number is
, so the largest number is
.
Solution 2
Use the answers to attempt to "reverse engineer" an appropriate pair of numbers. Looking at option , guess that one of the numbers is
. If the sum of two numbers is
and one is
, then other must be
. The product of those numbers is
, which is the second condition of the problem, so our number are
and
.
However, is the smaller of the two numbers, so the answer is
or
.
Solution 3
We go through the divisor pairs of to see which two numbers sum to
. 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
, namely
and
,
and
,
and
, and
and
. The only pair of numbers that sums to
is
and
. The larger number is
, so the answer is
.
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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