Difference between revisions of "2005 AIME II Problems/Problem 13"
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==Solution 2== | ==Solution 2== | ||
− | 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>418</math>. | + | 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>. |
-David Camacho | -David Camacho | ||
==Solution 3== | ==Solution 3== | ||
− | 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>. | + | 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>. |
~ Alcumus (Solution) | ~ Alcumus (Solution) |
Revision as of 13:55, 12 August 2019
Contents
[hide]Problem
Let be a polynomial with integer coefficients that satisfies and Given that has two distinct integer solutions and find the product
Solution
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 .
Solution 2
We know that so has two distinct solutions so is at least quadratic. Let us first try this problem out as if is a quadratic polynomial. Thus because where are all integers. Thus where are all integers. We know that or and or . By doing we obtain that or or . Thus . Now we know that , we have or which makes . Thus . By Vieta's formulas, we know that the sum of the roots() is equal to 41 and the product of the roots() is equal to . Because the roots are integers has to be an integer, so . Thus the product of the roots is equal to one of the following: . Testing every potential product of the roots, we find out that the only product that can have divisors that sum up to is .
-David Camacho
Solution 3
We have . Using the property that whenever and are distinct integers, we get and Since and , we must have We look for two divisors of that differ by ; we find that and satisfies these conditions. Therefore, either , giving , or , giving . From this, we conclude that and that .
~ Alcumus (Solution)
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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