Difference between revisions of "2005 AIME II Problems/Problem 13"
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== Solution == | == Solution == | ||
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+ | === Solution 1 === | ||
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_2 - 24</math> must divide <math>Q(n_2) - Q(24) = 10</math>, so <math>n_2</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 <math>19</math> and the other is <math>22</math> and their product is <math>19 \cdot 22 = \boxed{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_2 - 24</math> must divide <math>Q(n_2) - Q(24) = 10</math>, so <math>n_2</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 <math>19</math> and the other is <math>22</math> and their product is <math>19 \cdot 22 = \boxed{418}</math>. | ||
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+ | === Solution 2 === | ||
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+ | As above, we define <math>Q(x)=P(x)-x+7</math>, noting that it has roots at <math>17</math> and <math>24</math>. Hence <math>P(x)-x+7=A(x-17)(x-24)</math>. In particular, this means that | ||
+ | <math>P(x)-x-3=A(x-17)(x-24)-10</math>. Therefore, <math>x=n_1,n_2</math> satisfy <math>A(x-17)(x-24)=10</math>, where <math>A</math>, <math>(x-17)</math>, and <math>(x-24)</math> are integers. This cannot occur if <math>x\le 17</math> or <math>x\ge 24</math> because the product <math>(x-17)(x-24)</math> will either be too large or not be a divisor of <math>10</math>. We find that <math>x=19</math> and <math>x=22</math> are the only values that allow <math>(x-17)(x-24)</math> to be a factor of <math>10</math>. Hence the answer is <math>19\cdot 22=\boxed{418}</math>. | ||
== See also == | == See also == |
Revision as of 18:09, 6 January 2016
Problem
Let be a polynomial with integer coefficients that satisfies and Given that has two distinct integer solutions and find the product
Solution
Solution 1
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 and the other is and their product is .
Solution 2
As above, we define , noting that it has roots at and . Hence . In particular, this means that . Therefore, satisfy , where , , and are integers. This cannot occur if or because the product will either be too large or not be a divisor of . We find that and are the only values that allow to be a factor of . Hence the answer 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 |
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