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Difference between revisions of "2005 AIME II Problems/Problem 13"

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==Solution 2==
 
==Solution 2==
  
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We know that <math>P(n)-(n+3)=0</math> so <math>P(n)</math> has two distinct solutions so <math>P(x)</math> is at least quadratic. Let us first try this problem out as if <math>P(x)</math> is a quadratic polynomial. Thus <math>P(n)-(n+3)= an^2+(b-1)n+(c-3)=0</math> because <math>P(n)=an^2+bn+c</math> where <math>a,b,c</math> are all integers. Thus <math>P(x)=ax^2+bx+c</math> where <math>a,b,c</math> are all integers. We know that <math>P(17)</math> or <math>289a+17b+c=10</math> and <math>P(24)</math> or <math>576a+24b+c=17</math>. By doing <math>P(24)-P(17)</math> we obtain that <math>287a+7b=7</math> or <math>41a+b=1</math> or <math>-41a=b-1</math>. Thus <math>P(n)=an^2- (41a)n+(c-3)=0</math>.  Now we know that <math>b=-41a+1</math>, we have <math>289a+17(-41a+1)+c=10</math> or <math>408a=7+c</math> which makes <math>408a-10=c-3</math>. Thus <math>P(n)=an^2-(41a)n+(408a-10)=0</math>. By Vieta's formulas, we know that the sum of the roots(<math>n</math>) is equal to 41 and the product of the roots(<math>n</math>) is equal to <math>408-/frac[10][a]</math>. Because the roots are integers <math>/frac[10][a]</math> has to be an integer, so <math>a=1,2,5,10,-1,-2,-5,-10</math>. Thus the product of the roots is equal to one of the following: <math>398,403,406,407,409,410,413,418</math>. Testing every potential product of the roots, we find out that the only product that can have divisors that sum up to 41 is 418.
+
We know that <math>P(n)-(n+3)=0</math> so <math>P(n)</math> has two distinct solutions so <math>P(x)</math> is at least quadratic. Let us first try this problem out as if <math>P(x)</math> is a quadratic polynomial. Thus <math>P(n)-(n+3)= an^2+(b-1)n+(c-3)=0</math> because <math>P(n)=an^2+bn+c</math> where <math>a,b,c</math> are all integers. Thus <math>P(x)=ax^2+bx+c</math> where <math>a,b,c</math> are all integers. We know that <math>P(17)</math> or <math>289a+17b+c=10</math> and <math>P(24)</math> or <math>576a+24b+c=17</math>. By doing <math>P(24)-P(17)</math> we obtain that <math>287a+7b=7</math> or <math>41a+b=1</math> or <math>-41a=b-1</math>. Thus <math>P(n)=an^2- (41a)n+(c-3)=0</math>.  Now we know that <math>b=-41a+1</math>, we have <math>289a+17(-41a+1)+c=10</math> or <math>408a=7+c</math> which makes <math>408a-10=c-3</math>. Thus <math>P(n)=an^2-(41a)n+(408a-10)=0</math>. By Vieta's formulas, we know that the sum of the roots(<math>n</math>) is equal to 41 and the product of the roots(<math>n</math>) is equal to <math>408-/frac{10}{a}</math>. Because the roots are integers <math>/frac{10}{a}</math> has to be an integer, so <math>a=1,2,5,10,-1,-2,-5,-10</math>. Thus the product of the roots is equal to one of the following: <math>398,403,406,407,409,410,413,418</math>. Testing every potential product of the roots, we find out that the only product that can have divisors that sum up to 41 is 418.
  
 
== See also ==
 
== See also ==

Revision as of 23:44, 7 August 2018

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

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 $19\cdot 22=\boxed{418}$.


Solution 2

We know that $P(n)-(n+3)=0$ so $P(n)$ has two distinct solutions so $P(x)$ is at least quadratic. Let us first try this problem out as if $P(x)$ is a quadratic polynomial. Thus $P(n)-(n+3)= an^2+(b-1)n+(c-3)=0$ because $P(n)=an^2+bn+c$ where $a,b,c$ are all integers. Thus $P(x)=ax^2+bx+c$ where $a,b,c$ are all integers. We know that $P(17)$ or $289a+17b+c=10$ and $P(24)$ or $576a+24b+c=17$. By doing $P(24)-P(17)$ we obtain that $287a+7b=7$ or $41a+b=1$ or $-41a=b-1$. Thus $P(n)=an^2- (41a)n+(c-3)=0$. Now we know that $b=-41a+1$, we have $289a+17(-41a+1)+c=10$ or $408a=7+c$ which makes $408a-10=c-3$. Thus $P(n)=an^2-(41a)n+(408a-10)=0$. By Vieta's formulas, we know that the sum of the roots($n$) is equal to 41 and the product of the roots($n$) is equal to $408-/frac{10}{a}$. Because the roots are integers $/frac{10}{a}$ has to be an integer, so $a=1,2,5,10,-1,-2,-5,-10$. Thus the product of the roots is equal to one of the following: $398,403,406,407,409,410,413,418$. Testing every potential product of the roots, we find out that the only product that can have divisors that sum up to 41 is 418.

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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