Difference between revisions of "2005 AIME II Problems/Problem 13"
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== Solution == | == Solution == | ||
− | + | 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>. | ||
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+ | |||
+ | ==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>\boxed{418}</math>. | ||
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+ | -David Camacho | ||
+ | |||
+ | ==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> and that <math>n_1\cdot n_2=\boxed{418}</math>. | ||
+ | |||
+ | ~ Alcumus (Solution) | ||
== See also == | == See also == | ||
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[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 02:14, 1 June 2024
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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