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 | ||
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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. | ||
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Thus, the answer is: <cmath>\boxed{600}.</cmath> | Thus, the answer is: <cmath>\boxed{600}.</cmath> | ||
+ | |||
+ | ~Minor edit by Yiyj1 | ||
==Solution 2== | ==Solution 2== | ||
− | Similar to the previous | + | 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)== | ==Solution 3 (less complicated)== | ||
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<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 | ||
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Solution provided by- Yonglao | Solution provided by- Yonglao | ||
+ | |||
+ | |||
+ | Remark : This solution works because | ||
+ | no distinct <math>(a,b), (c,d)</math> exist such that <math>a+b=c+d,ab=cd</math> | ||
==Solution 5== | ==Solution 5== | ||
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==Solution 6(simple)== | ==Solution 6(simple)== | ||
− | By | + | 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 | -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 7 (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 8 (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 01:26, 5 May 2023
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
Remark : This solution works because
no distinct exist such that
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 7 (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 8 (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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