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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
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<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 17:09, 6 January 2016

Problem

Let $P(x)$ be a polynomial with integer coefficients that satisfies $P(17)=10$ and $P(24)=17.$ Given that $P(n)=n+3$ has two distinct integer solutions $n_1$ and $n_2,$ find the product $n_1\cdot n_2.$

Solution

Solution 1

Define the polynomial $Q(x) = P(x) - x + 7$. By the givens, $Q(17) = 10 - 17 + 7 = 0$, $Q(24) = 17 - 24 + 7 = 0$, $Q(n_1) = n_1 + 3 - n_1 + 7 = 10$ and $Q(n_2) = n_2 + 3 - n_2 + 7 = 10$. Note that for any polynomial $P(x)$ with integer coefficients and any integers $a, b$ we have $a - b$ divides $P(a)-P(b)$. So $n_1 - 17$ divides $Q(n_1) - Q(17) = 10$, and so $n_1 - 17$ must be one of the eight numbers $\pm1, \pm2, \pm5, \pm10$ and so $n_1$ must be one of the numbers $7, 12, 15, 16, 18, 19, 22$ or $27$. Similarly, $n_2 - 24$ must divide $Q(n_2) - Q(24) = 10$, so $n_2$ must be one of the eight numbers $14, 19, 22, 23, 25, 26, 29$ or $34$. Thus, $n_1$ must be either 19 or 22. Since $n_2$ obeys the same conditions and $n_1$ and $n_2$ are different, one of them is $19$ and the other is $22$ and their product is $19 \cdot 22 = \boxed{418}$.

Solution 2

As above, we define $Q(x)=P(x)-x+7$, noting that it has roots at $17$ and $24$. Hence $P(x)-x+7=A(x-17)(x-24)$. In particular, this means that $P(x)-x-3=A(x-17)(x-24)-10$. Therefore, $x=n_1,n_2$ satisfy $A(x-17)(x-24)=10$, where $A$, $(x-17)$, and $(x-24)$ are integers. This cannot occur if $x\le 17$ or $x\ge 24$ because the product $(x-17)(x-24)$ will either be too large or not be a divisor of $10$. We find that $x=19$ and $x=22$ are the only values that allow $(x-17)(x-24)$ to be a factor of $10$. Hence the answer is $19\cdot 22=\boxed{418}$.

See also

2005 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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All AIME Problems and Solutions

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