Difference between revisions of "2018 AIME I Problems/Problem 1"
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Let <math>S</math> be the number of ordered pairs of integers <math>(a,b)</math> with <math>1 \leq a \leq 100</math> and <math>b \geq 0</math> such that the polynomial <math>x^2+ax+b</math> can be factored into the product of two (not necessarily distinct) linear factors with integer coefficients. Find the remainder when <math>S</math> is divided by <math>1000</math>. | Let <math>S</math> be the number of ordered pairs of integers <math>(a,b)</math> with <math>1 \leq a \leq 100</math> and <math>b \geq 0</math> such that the polynomial <math>x^2+ax+b</math> can be factored into the product of two (not necessarily distinct) linear factors with integer coefficients. Find the remainder when <math>S</math> is divided by <math>1000</math>. | ||
− | ==Solution== | + | ==Solution 1== |
− | + | Let the linear factors be <math>(x+c)(x+d)</math>. | |
− | Then, | + | Then, <math>a=c+d</math> and <math>b=cd</math>. |
We know that <math>1\le a\le 100</math> and <math>b\ge 0</math>, so <math>c</math> and <math>d</math> both have to be non-negative | We know that <math>1\le a\le 100</math> and <math>b\ge 0</math>, so <math>c</math> and <math>d</math> both have to be non-negative | ||
Line 14: | Line 14: | ||
Also, <math>a</math> cannot be greater than <math>100</math>, so <math>c+d</math> must be less than or equal to <math>100</math>. | Also, <math>a</math> cannot be greater than <math>100</math>, so <math>c+d</math> must be less than or equal to <math>100</math>. | ||
− | Essentially, if we plot the solutions, we get a triangle on the coordinate plane with vertices <math>(0,0), (0, 100),</math> and <math>(100,0)</math>. Remember that <math>(0,0)</math> does not work, so there is a square with top right corner <math>(1,1)</math>. | + | Essentially, if we plot the solutions, we get a triangle on the coordinate plane with vertices <math>(0,0), (0, 100),</math> and <math>(100,0)</math>. Remember that <math>(0,0)</math> does not work, so there is a square with the top right corner <math>(1,1)</math>. |
− | Note that <math>c</math> and <math>d</math> are interchangeable | + | Note that <math>c</math> and <math>d</math> are interchangeable since they end up as <math>a</math> and <math>b</math> in the end anyways. Thus, we simply draw a line from <math>(1,1)</math> to <math>(50,50)</math>, designating one of the halves as our solution (since the other side is simply the coordinates flipped). |
We note that the pattern from <math>(1,1)</math> to <math>(50,50)</math> is <math>2+3+4+\dots+51</math> solutions and from <math>(51, 49)</math> to <math>(100,0)</math> is <math>50+49+48+\dots+1</math> solutions, since we can decrease the <math>y</math>-value by <math>1</math> until <math>0</math> for each coordinate. | We note that the pattern from <math>(1,1)</math> to <math>(50,50)</math> is <math>2+3+4+\dots+51</math> solutions and from <math>(51, 49)</math> to <math>(100,0)</math> is <math>50+49+48+\dots+1</math> solutions, since we can decrease the <math>y</math>-value by <math>1</math> until <math>0</math> for each coordinate. | ||
Line 25: | Line 25: | ||
Thus, the answer is: <cmath>\boxed{600}.</cmath> | Thus, the answer is: <cmath>\boxed{600}.</cmath> | ||
− | + | ~Minor edit by Yiyj1 | |
− | + | ==Solution 2== | |
− | ==Solution | + | Similar to the previous solution, we plot the triangle and cut it in half. Then, we find the number of boundary points, which is <math>101+51+51-3</math>, or just <math>200</math>. Using Pick's theorem, we know that the area of the half-triangle, which is <math>2500</math>, is just <math>I+100-1</math>. That means that there are <math>2401</math> interior points, plus <math>200</math> boundary points, which is <math>2601</math>. However, <math>(0,0)</math> does not work, so the answer is <cmath>\boxed{600}.</cmath> |
+ | |||
+ | ==Solution 3 (less complicated)== | ||
Notice that for <math>x^2+ax+b</math> to be true, for every <math>a</math>, <math>b</math> will always be the product of the possibilities of how to add two integers to <math>a</math>. For example, if <math>a=3</math>, <math>b</math> will be the product of <math>(3,0)</math> and <math>(2,1)</math>, as those two sets are the only possibilities of adding two integers to <math>a</math>. Note that order does not matter. If we just do some simple casework, we find out that: | Notice that for <math>x^2+ax+b</math> to be true, for every <math>a</math>, <math>b</math> will always be the product of the possibilities of how to add two integers to <math>a</math>. For example, if <math>a=3</math>, <math>b</math> will be the product of <math>(3,0)</math> and <math>(2,1)</math>, as those two sets are the only possibilities of adding two integers to <math>a</math>. Note that order does not matter. If we just do some simple casework, we find out that: | ||
Line 42: | Line 44: | ||
Solution by IronicNinja~ | Solution by IronicNinja~ | ||
− | ==Solution | + | ==Solution 4== |
Let's write the linear factors as <math>(x+n)(x+m)</math>. | Let's write the linear factors as <math>(x+n)(x+m)</math>. | ||
Line 59: | Line 61: | ||
<math>1, 2, 2, 3, 3, 4, 4…</math> | <math>1, 2, 2, 3, 3, 4, 4…</math> | ||
+ | === Ending 1 === | ||
+ | We see the pattern <math>1, 2; 2, 3; 3, 4; ...</math>. There are 50 pairs of <math>i, i+1</math> in this pattern, and each pair sums to <math>2i+1</math>. So the pattern condenses to <math>3, 5, 7, ...</math> for 50 terms. This is just <math>1+3+5+...</math> for 51 terms, minus <math>1</math>, or <math>51^2-1=2601-1=2600\implies\boxed{600}</math>. | ||
+ | |||
+ | ~ Firebolt360 | ||
+ | |||
+ | === Ending 2 === | ||
The following link is the URL to the graph I drew showing the relationship between a-values and b-values | The following link is the URL to the graph I drew showing the relationship between a-values and b-values | ||
http://artofproblemsolving.com/wiki/index.php?title=File:Screen_Shot_2018-04-30_at_8.15.00_PM.png#file | http://artofproblemsolving.com/wiki/index.php?title=File:Screen_Shot_2018-04-30_at_8.15.00_PM.png#file | ||
+ | [[File:Screen Shot 2018-04-30 at 8.15.00 PM.png|thumb|240px]] | ||
The pattern continues until <math>a=100</math>, and in total, there are <math>49</math> pairs of a-value with the same amount of b-values. The two lone a-values without a pair are, the (<math>a=1</math>, amount of b-values=1) in the beginning, and (<math>a=100</math>, amount of b-values=51) in the end. | The pattern continues until <math>a=100</math>, and in total, there are <math>49</math> pairs of a-value with the same amount of b-values. The two lone a-values without a pair are, the (<math>a=1</math>, amount of b-values=1) in the beginning, and (<math>a=100</math>, amount of b-values=51) in the end. | ||
Line 71: | Line 80: | ||
Solution provided by- Yonglao | Solution provided by- Yonglao | ||
+ | Slightly modified by Afly | ||
+ | |||
+ | |||
+ | Remark : This solution works because | ||
+ | no distinct <math>(a,b), (c,d)</math> exist such that <math>a+b=c+d,ab=cd</math> unless <math>a = d</math> and <math>b = c</math> | ||
+ | |||
+ | ==Solution 5== | ||
+ | |||
+ | Let's say that the quadratic <math>x^2 + ax + b</math> can be factored into <math>(x+c)(x+d)</math> where <math>c</math> and <math>d</math> are non-negative numbers. We can't have both of them zero because <math>a</math> would not be within bounds. Also, <math>c+d \leq 100</math>. Assume that <math>c < d</math>. <math>d</math> can be written as <math>c + x</math> where <math>x \geq 0</math>. Therefore, <math>c + d = 2c + x \leq 100</math>. To find the amount of ordered pairs, we must consider how many values of <math>x</math> are possible for each value of <math>c</math>. The amount of possible values of <math>x</math> is given by <math>100 - 2c + 1</math>. The <math>+1</math> is the case where <math>c = d</math>. We don't include the case where <math>c = d = 0</math>, so we must subtract a case from our total. The amount of ordered pairs of <math>(a,b)</math> is: | ||
+ | <cmath>\left(\sum_{c=0}^{50} (100 - 2c + 1)\right) - 1</cmath> | ||
+ | This is an arithmetic progression. <cmath>\frac{(101 + 1)(51)}{2} - 1 = 2601 - 1 = 2600</cmath> | ||
+ | When divided by <math>1000</math>, it gives the remainder <math>\boxed{600}</math> | ||
+ | |||
+ | ~Zeric Hang | ||
+ | |||
+ | ==Solution 6(simple)== | ||
+ | By Vietas, the sum of the roots is <math>-a</math> and the product is <math>b</math>. Therefore, both roots are nonpositive. For each value of <math>a</math> from <math>1</math> to <math>100</math>, the number of <math>b</math> values is the number of ways to sum two numbers between <math>0</math> and <math>a-1</math> inclusive to <math>a</math>. This is just <math>1 + 2 + 2 + 3 + 3 +... 50 + 50 + 51 = 2600</math>. Thus, the answer is <math>\boxed{600}</math> | ||
+ | |||
+ | -bron jiang | ||
+ | |||
+ | ==Solution 7 (less room for error)== | ||
+ | Similar to solution 1 we plot the triangle and half it. From dividing the triangle in half we are removing the other half of answers that are just flipped coordinates. We notice that we can measure the length of the longest side of the half triangle which is just from <math>0</math> to <math>100</math>, so the number of points on that line is is <math>101</math>. The next row has length <math>99</math>, the one after that has length <math>97</math>, and so on. We simply add this arithmetic series of odd integers <math>101+99+97+...+1</math>. | ||
+ | This is <cmath>\frac{(101+1)(51)}{2} = 2601</cmath> | ||
+ | Or you can notice that this is the sum of the first <math>51</math> odd terms, which is just <math>51^2 = 2601</math>. | ||
+ | However, <math>(0,0)</math> is the singular coordinate that does not work, so the answer is <math>(2601-1)\bmod {1000}\equiv\boxed{600}</math> | ||
+ | |||
+ | Solution by Damian Kim~ | ||
+ | |||
+ | ==Solution 8 (Sticks and stones)== | ||
+ | Letting <math>-r</math> and <math>-q</math> be the roots, we must find the number of unordered pairs <math>(r,q)</math> such that <math>1 \le r+q \le 100.</math> To do this we define three boxes: <math>r, q,</math> and <math>t</math> for "trash." For example, if we had <math>t=77,</math> we would have any arrangement of <math>r</math> and <math>q</math> summing to <math>23.</math> Note that we cannot have <math>t=100,</math> subtracting one from our total. By stars and bars, we have that the number of ordered pairs <math>(r,q)</math> is <math>{102 \choose 2} - 1 = 5150.</math> However, we are not done: we have double-counted cases such as <math>(1,2)</math> and <math>(2,1)</math> but not cases such as <math>(4,4).</math> Thus our answer will be <math>\frac{5150 - 50}{2} + 50 \pmod {100} = \boxed{600}.</math> | ||
+ | ~ab2024 | ||
+ | |||
+ | ==Solution 9 (Official MAA)== | ||
+ | The factoring condition is equivalent to the discriminant <math>a^2-4b</math> being equal to <math>c^2</math> for some integer <math>c.</math> Because <math>b\ge 0,</math> the equation <math>4b=(a-c)(a+c)</math> shows that the existence of such a <math>b</math> is equivalent to <math>a\equiv c\pmod 2</math> with <math>0<c\le a.</math> Thus the number of ordered pairs is <cmath>S=\sum_{a=1}^{100}\left\lceil\frac{a+1}{2}\right\rceil=2600.</cmath> The requested remainder is <math>600.</math> | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=WVtbD8x9fCM | ||
+ | ~Shreyas S | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2018|n=I|before=First Problem|num-a=2}} | {{AIME box|year=2018|n=I|before=First Problem|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 20:57, 20 June 2024
Contents
Problem 1
Let be the number of ordered pairs of integers
with
and
such that the polynomial
can be factored into the product of two (not necessarily distinct) linear factors with integer coefficients. Find the remainder when
is divided by
.
Solution 1
Let the linear factors be .
Then, and
.
We know that and
, so
and
both have to be non-negative
However, cannot be
, so at least one of
and
must be greater than
, ie positive.
Also, cannot be greater than
, so
must be less than or equal to
.
Essentially, if we plot the solutions, we get a triangle on the coordinate plane with vertices and
. Remember that
does not work, so there is a square with the top right corner
.
Note that and
are interchangeable since they end up as
and
in the end anyways. Thus, we simply draw a line from
to
, designating one of the halves as our solution (since the other side is simply the coordinates flipped).
We note that the pattern from to
is
solutions and from
to
is
solutions, since we can decrease the
-value by
until
for each coordinate.
Adding up gives
This gives us
, and
Thus, the answer is:
~Minor edit by Yiyj1
Solution 2
Similar to the previous solution, we plot the triangle and cut it in half. Then, we find the number of boundary points, which is , or just
. Using Pick's theorem, we know that the area of the half-triangle, which is
, is just
. That means that there are
interior points, plus
boundary points, which is
. However,
does not work, so the answer is
Solution 3 (less complicated)
Notice that for to be true, for every
,
will always be the product of the possibilities of how to add two integers to
. For example, if
,
will be the product of
and
, as those two sets are the only possibilities of adding two integers to
. Note that order does not matter. If we just do some simple casework, we find out that:
if is odd, there will always be
possibilities of adding two integers to
.
if is even, there will always be
possibilities of adding two integers to
.
Using the casework, we have possibilities. This will mean that the answer is
possibilities.
Thus, our solution is .
Solution by IronicNinja~
Solution 4
Let's write the linear factors as .
Then we can write them as: .
or
has to be a positive integer as a cannot be 0.
has to be between
and
, as a cannot be over
.
Excluding , we can see there is always a pair of
a-values for a certain amount of b-values.
For instance, and
both have
b-values.
and
both have
b-values.
We notice the pattern of the number of b-values in relation to the a-values:
Ending 1
We see the pattern . There are 50 pairs of
in this pattern, and each pair sums to
. So the pattern condenses to
for 50 terms. This is just
for 51 terms, minus
, or
.
~ Firebolt360
Ending 2
The following link is the URL to the graph I drew showing the relationship between a-values and b-values http://artofproblemsolving.com/wiki/index.php?title=File:Screen_Shot_2018-04-30_at_8.15.00_PM.png#file
The pattern continues until , and in total, there are
pairs of a-value with the same amount of b-values. The two lone a-values without a pair are, the (
, amount of b-values=1) in the beginning, and (
, amount of b-values=51) in the end.
Then, we add numbers from the opposite ends of the spectrum, and quickly notice that there are pairs each with a sum of
.
gives
ordered pairs:
When divided by , it gives the remainder
, the answer.
Solution provided by- Yonglao Slightly modified by Afly
Remark : This solution works because
no distinct exist such that
unless
and
Solution 5
Let's say that the quadratic can be factored into
where
and
are non-negative numbers. We can't have both of them zero because
would not be within bounds. Also,
. Assume that
.
can be written as
where
. Therefore,
. To find the amount of ordered pairs, we must consider how many values of
are possible for each value of
. The amount of possible values of
is given by
. The
is the case where
. We don't include the case where
, so we must subtract a case from our total. The amount of ordered pairs of
is:
This is an arithmetic progression.
When divided by
, it gives the remainder
~Zeric Hang
Solution 6(simple)
By Vietas, the sum of the roots is and the product is
. Therefore, both roots are nonpositive. For each value of
from
to
, the number of
values is the number of ways to sum two numbers between
and
inclusive to
. This is just
. Thus, the answer is
-bron jiang
Solution 7 (less room for error)
Similar to solution 1 we plot the triangle and half it. From dividing the triangle in half we are removing the other half of answers that are just flipped coordinates. We notice that we can measure the length of the longest side of the half triangle which is just from to
, so the number of points on that line is is
. The next row has length
, the one after that has length
, and so on. We simply add this arithmetic series of odd integers
.
This is
Or you can notice that this is the sum of the first
odd terms, which is just
.
However,
is the singular coordinate that does not work, so the answer is
Solution by Damian Kim~
Solution 8 (Sticks and stones)
Letting and
be the roots, we must find the number of unordered pairs
such that
To do this we define three boxes:
and
for "trash." For example, if we had
we would have any arrangement of
and
summing to
Note that we cannot have
subtracting one from our total. By stars and bars, we have that the number of ordered pairs
is
However, we are not done: we have double-counted cases such as
and
but not cases such as
Thus our answer will be
~ab2024
Solution 9 (Official MAA)
The factoring condition is equivalent to the discriminant being equal to
for some integer
Because
the equation
shows that the existence of such a
is equivalent to
with
Thus the number of ordered pairs is
The requested remainder is
Video Solution
https://www.youtube.com/watch?v=WVtbD8x9fCM ~Shreyas S
See Also
2018 AIME I (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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