Difference between revisions of "2022 AMC 12B Problems/Problem 4"
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We list out the factors of <math>36</math> as follows: <cmath>\{1,2,3,4,6,9,12,18,36\}</cmath> | We list out the factors of <math>36</math> as follows: <cmath>\{1,2,3,4,6,9,12,18,36\}</cmath> | ||
Then, we notice that there are exactly 4 pairings of distinct integers that multiply to <math>36</math> (everything excluding <math>6</math>). | Then, we notice that there are exactly 4 pairings of distinct integers that multiply to <math>36</math> (everything excluding <math>6</math>). | ||
| + | Their corresponding <math>k = -(r_1 + r_2)</math> values are also distinct, which ensures we aren't double counting any pairs. | ||
However, we aren't done yet! <math>r_1</math> and <math>r_2</math> can be negative. Since the signs of <math>r_1</math> and <math>r_2</math> must match and being negative doesn't change the number of options, we simply double the number of positive pairings to arrive at <math>\boxed{\textbf{(B)} \ 8}</math> | However, we aren't done yet! <math>r_1</math> and <math>r_2</math> can be negative. Since the signs of <math>r_1</math> and <math>r_2</math> must match and being negative doesn't change the number of options, we simply double the number of positive pairings to arrive at <math>\boxed{\textbf{(B)} \ 8}</math> | ||
| + | |||
~ <math>\color{magenta} zoomanTV</math> | ~ <math>\color{magenta} zoomanTV</math> | ||
Revision as of 17:13, 17 November 2022
Problem 4
For how many values of the constant 
will the polynomial 
have two distinct integer roots?
Solution
Using Vieta's, denote the two distinct integer roots as 
. Then
![\[r_1 \cdot r_2 = 36\]](http://latex.artofproblemsolving.com/b/c/5/bc527012cd5bebfdbaaeb19aa116963a8b55b1d3.png)
We list out the factors of 
as follows: ![\[\{1,2,3,4,6,9,12,18,36\}\]](http://latex.artofproblemsolving.com/b/7/a/b7a945e59c48bb0b42ee9eadf2d3d9509b46cccd.png)
Then, we notice that there are exactly 4 pairings of distinct integers that multiply to (everything excluding 
).
Their corresponding 
values are also distinct, which ensures we aren't double counting any pairs.
However, we aren't done yet! 
and 
can be negative. Since the signs of and must match and being negative doesn't change the number of options, we simply double the number of positive pairings to arrive at 
~ 
See Also
| 2022 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 3 |
Followed by Problem 5 |
| 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 AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
