Difference between revisions of "2005 AIME II Problems/Problem 13"
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== Solution == | == Solution == | ||
| − | Define the [[polynomial]] <math>Q(x) = P(x) - x + 7</math>. By the givens, <math>Q(17) = 10 - 17 + 7 = 0</math>, <math>Q(24) = 17 - 24 + 7 = 0</math>, <math>Q(n_1) = n_1 + 3 - n_1 + 7 = 10</math> and <math>Q(n_2) = n_2 + 3 - n_2 + 7 = 10</math>. Note that for any polynomial <math>P(x)</math> with [[integer]] [[coefficient]]s and any integers <math>a, b</math> we have <math>a - b</math> [[divisor|divides]] <math>P(a)-P(b)</math>. So <math>n_1 - 17</math> divides <math>Q(n_1) - Q(17) = 10</math>, and so <math>n_1 - 17</math> must be one of the eight numbers <math>\pm1, \pm2, \pm5, \pm10</math> and so <math>n_1</math> must be one of the numbers <math>7, 12, 15, 16, 18, 19, 22</math> or <math>27</math>. Similarly, <math>n_1 - 24</math> must divide <math>Q(n_1) - Q(24) = 10</math>, so <math>n_1</math> must be one of the eight numbers <math>14, 19, 22, 23, 25, 26, 29</math> or <math>34</math>. Thus, <math>n_1</math> must be either 19 or 22. Since <math>n_2</math> obeys the same conditions and <math>n_1</math> and <math>n_2</math> are different, one of them is 19 and the other is 22 and their product is <math>19 \cdot 22 = 418</math>. | + | Define the [[polynomial]] <math>Q(x) = P(x) - x + 7</math>. By the givens, <math>Q(17) = 10 - 17 + 7 = 0</math>, <math>Q(24) = 17 - 24 + 7 = 0</math>, <math>Q(n_1) = n_1 + 3 - n_1 + 7 = 10</math> and <math>Q(n_2) = n_2 + 3 - n_2 + 7 = 10</math>. Note that for any polynomial <math>P(x)</math> with [[integer]] [[coefficient]]s and any integers <math>a, b</math> we have <math>a - b</math> [[divisor|divides]] <math>P(a)-P(b)</math>. So <math>n_1 - 17</math> divides <math>Q(n_1) - Q(17) = 10</math>, and so <math>n_1 - 17</math> must be one of the eight numbers <math>\pm1, \pm2, \pm5, \pm10</math> and so <math>n_1</math> must be one of the numbers <math>7, 12, 15, 16, 18, 19, 22</math> or <math>27</math>. Similarly, <math>n_1 - 24</math> must divide <math>Q(n_1) - Q(24) = 10</math>, so <math>n_1</math> must be one of the eight numbers <math>14, 19, 22, 23, 25, 26, 29</math> or <math>34</math>. Thus, <math>n_1</math> must be either 19 or 22. Since <math>n_2</math> obeys the same conditions and <math>n_1</math> and <math>n_2</math> are different, one of them is 19 and the other is 22 and their product is <math>19 \cdot 22 = \boxed{418}</math>. |
== See also == | == See also == | ||
Revision as of 19:18, 16 March 2014
Problem
Let 
be a polynomial with integer coefficients that satisfies 
and 
Given that 
has two distinct integer solutions 
and 
find the product 
Solution
Define the polynomial 
. By the givens, 
, 
, 
and 
. Note that for any polynomial with integer coefficients and any integers 
we have 
divides 
. So 
divides 
, and so must be one of the eight numbers 
and so must be one of the numbers 
or 
. Similarly, 
must divide 
, so must be one of the eight numbers 
or 
. Thus, must be either 19 or 22. Since 
obeys the same conditions and and are different, one of them is 19 and the other is 22 and their product is 
.
See also
| 2005 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
