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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% so P(n) has two distinct solutions so %P(x)% is at least quadratic. Let us first try this problem out as if </math>P(x)<math> is a quadratic polynomial. Thus </math>P(x)=ax^2+bx+c<math> where </math>a,b,c$ are all integers.
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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 <math>41</math> is <math>\boxed{418}</math>.
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-David Camacho
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==Solution 3==
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We have <math>P(n_1) = n_1+3</math>. Using the property that <math>a - b \mid P(a) - P(b)</math> whenever <math>a</math> and <math>b</math> are distinct integers, we get <cmath>n_1 - 17 \mid P(n_1) - P(17) = (n_1+3) - 10 = n_1 - 7,</cmath>and <cmath>n_1 - 24 \mid P(n_1) - P(24) = (n_1+3)-17=n_1-14.</cmath>Since <math>n_1 - 7 = 10 + (n_1-17)</math> and <math>n_1-14 = 10 + (n_1-24)</math>, we must have <cmath>n_1 - 17 \mid 10 \; \text{and} \; n_1-24 \mid 10.</cmath>We look for two divisors of <math>10</math> that differ by <math>7</math>; we find that <math>\{2, -5\}</math> and <math>\{5, -2\}</math> satisfies these conditions. Therefore, either <math>n_1 - 24 = -5</math>, giving <math>n_1 = 19</math>, or <math>n_1 - 24 = -2</math>, giving <math>n_1 = 22</math>. From this, we conclude that <math>n_1, n_2 = \boxed{19, 22}</math> and that <math>n_1\cdot n_2=\boxed{418}</math>.
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~ Alcumus (Solution)
  
 
== See also ==
 
== See also ==

Latest revision as of 01:14, 1 June 2024

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 $\boxed{418}$.

-David Camacho

Solution 3

We have $P(n_1) = n_1+3$. Using the property that $a - b \mid P(a) - P(b)$ whenever $a$ and $b$ are distinct integers, we get \[n_1 - 17 \mid P(n_1) - P(17) = (n_1+3) - 10 = n_1 - 7,\]and \[n_1 - 24 \mid P(n_1) - P(24) = (n_1+3)-17=n_1-14.\]Since $n_1 - 7 = 10 + (n_1-17)$ and $n_1-14 = 10 + (n_1-24)$, we must have \[n_1 - 17 \mid 10 \; \text{and} \; n_1-24 \mid 10.\]We look for two divisors of $10$ that differ by $7$; we find that $\{2, -5\}$ and $\{5, -2\}$ satisfies these conditions. Therefore, either $n_1 - 24 = -5$, giving $n_1 = 19$, or $n_1 - 24 = -2$, giving $n_1 = 22$. From this, we conclude that $n_1, n_2 = \boxed{19, 22}$ and that $n_1\cdot n_2=\boxed{418}$.

~ Alcumus (Solution)

See also

2005 AIME II (Problems • Answer Key • Resources)
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Problem 12 		Followed by
Problem 14
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