Difference between revisions of "2020 AMC 10A Problems/Problem 17"
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\left(4^2,5^2\right) & 96 & \\ [0.5ex] | \left(4^2,5^2\right) & 96 & \\ [0.5ex] | ||
\left(5^2,6^2\right) & 95 & \checkmark \\ [0.5ex] | \left(5^2,6^2\right) & 95 & \checkmark \\ [0.5ex] | ||
− | \left(6^2,7^2\right) & 94 & \\ | + | \left(6^2,7^2\right) & 94 & \\ |
− | \ | + | \vdots & \vdots & \vdots \\ [0.75ex] |
\left(99^2,100^2\right) & 1 & \checkmark \\ [0.5ex] | \left(99^2,100^2\right) & 1 & \checkmark \\ [0.5ex] | ||
\left(100^2,\infty\right) & 0 & \\ [0.5ex] | \left(100^2,\infty\right) & 0 & \\ [0.5ex] | ||
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~MRENTHUSIASM (Reformatting) | ~MRENTHUSIASM (Reformatting) | ||
− | == Solution 2 == | + | == Solution 2 (Casework) == |
− | Notice that <math>P(x)</math> is nonpositive when <math>x</math> is between <math>100^2</math> and <math>99^2 | + | Notice that <math>P(x)</math> is nonpositive when <math>x</math> is between <math>100^2</math> and <math>99^2, 98^2</math> and <math>97^2, \cdots</math> , <math>2^2</math> and <math>1^2</math> (inclusive), because there are an odd number of negatives, which means that the number of values equals <cmath>((100+99)(100-99) + 1) + ((98+97)(98-97)+1) + \cdots + ((2+1)(2-1)+1).</cmath> |
− | This reduces to <cmath>200 + 196 + 192 + \ | + | This reduces to <cmath>200 + 196 + 192 + \cdots + 4 = 4(1+2+\cdots + 50) = 4 \cdot\frac{50 \cdot 51}{2} = \boxed{\textbf{(E) } 5100}.</cmath> |
~Zeric | ~Zeric | ||
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We know that <math>P(x)</math> is a <math>100</math>-degree function with a positive leading coefficient. That is, <math>P(x)=x^{100}+ax^{99}+bx^{98}+...+\text{(constant)}</math>. | We know that <math>P(x)</math> is a <math>100</math>-degree function with a positive leading coefficient. That is, <math>P(x)=x^{100}+ax^{99}+bx^{98}+...+\text{(constant)}</math>. | ||
− | Since the degree of <math>P(x)</math> is even, its end behaviors match. And since the leading coefficient is positive, we know that both ends approach <math>\infty</math> as <math>x</math> goes in either direction | + | Since the degree of <math>P(x)</math> is even, its end behaviors match. And since the leading coefficient is positive, we know that both ends approach <math>\infty</math> as <math>x</math> goes in either direction, from which <cmath>\lim_{x\to-\infty} P(x)=\lim_{x\to\infty} P(x)=\infty.</cmath> So the first time <math>P(x)</math> is going to be negative is when it intersects the <math>x</math>-axis at an <math>x</math>-intercept and it's going to dip below. This happens at <math>1^2</math>, which is the smallest intercept. |
− | |||
− | <cmath>\lim_{x\to-\infty} P(x)=\lim_{x\to\infty} P(x)=\infty</cmath> | ||
− | |||
− | So the first time <math>P(x)</math> is going to be negative is when it intersects the <math>x</math>-axis at an <math>x</math>-intercept and it's going to dip below. This happens at <math>1^2</math>, which is the smallest intercept. | ||
However, when it hits the next intercept, it's going to go back up again into positive territory, we know this happens at <math>2^2</math>. And when it hits <math>3^2</math>, it's going to dip back into negative territory. Clearly, this is going to continue to snake around the intercepts until <math>100^2</math>. | However, when it hits the next intercept, it's going to go back up again into positive territory, we know this happens at <math>2^2</math>. And when it hits <math>3^2</math>, it's going to dip back into negative territory. Clearly, this is going to continue to snake around the intercepts until <math>100^2</math>. | ||
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To get the amount of integers below and/or on the <math>x</math>-axis, we simply need to count the integers. For example, the amount of integers in between the <math>[1^2,2^2]</math> interval we got earlier, we subtract and add one. <math>(2^2-1^2+1)=4</math> integers, so there are four integers in this interval that produce a negative result. | To get the amount of integers below and/or on the <math>x</math>-axis, we simply need to count the integers. For example, the amount of integers in between the <math>[1^2,2^2]</math> interval we got earlier, we subtract and add one. <math>(2^2-1^2+1)=4</math> integers, so there are four integers in this interval that produce a negative result. | ||
− | Doing this with all of the other intervals, we have | + | Doing this with all of the other intervals, we have <cmath>(2^2-1^2+1)+(4^2-3^2+1)+\cdots+(100^2-99^2+1)=\boxed{\textbf{(E) } 5100}</cmath> from Solution 2's result. |
− | <math>(2^2-1^2+1)+ | + | ~quacker88 |
+ | |||
+ | == Solution 4 (Fast) == | ||
+ | We know <math>P(x) \leq 0</math> when an odd number of its factors are positive and negative. For example, to make the first factor positive, <math>x \in [1^2, 2^2]</math>. then there will be a even number of positive factors. We would do <math>2^2 - 1^2 + 1 (inclusive)</math> to find all integers that work. In short we can generalize too: | ||
+ | <cmath>\begin{align*} | ||
+ | x^2 - (x-1)^2 + 1 &= 2x \\ | ||
+ | x^2 - (x^2 - 2x + 1) + 1 &= 2x \\ | ||
+ | x^2 - x^2 + 2x - 1 + 1 &= 2x. \\ | ||
+ | \end{align*}</cmath> | ||
+ | But remember this only works when <math>x \in \{2, 4, 6, 8 \cdots 98, 100\}</math> because only then will there be a odd amount of positive and negative factors. So we can set <math>x = 2k</math>, for <math>k \in \{1, 2, 3, 4, \cdots 49, 50\}</math> Now we only have to solve: | ||
+ | <cmath>\sum_{k=1}^{k=50}2(2k) = 2\sum_{k = 1}^{k = 50}2k = 4\sum_{k = 1}^{k = 50}k = 4 \cdot \dfrac{(50)(51)}{2} = 2 \cdot (50)(51) = \boxed{\textbf{(E) } 5100}.</cmath> | ||
+ | ~Wiselion | ||
== Video Solutions == | == Video Solutions == |
Latest revision as of 10:50, 2 October 2023
Contents
Problem
Define How many integers
are there such that
?
Solution 1 (Casework)
We perform casework on
In this case, there are such integers
There are factors in
and we need an odd number of them to be negative. We construct the table below:
Note that there are
valid intervals of
We count the integers in these intervals:
In this case, there are
such integers
Together, the answer is
~PCChess (Solution)
~MRENTHUSIASM (Reformatting)
Solution 2 (Casework)
Notice that is nonpositive when
is between
and
and
,
and
(inclusive), because there are an odd number of negatives, which means that the number of values equals
This reduces to
~Zeric
~jesselan (Minor Edits)
Solution 3 (End Behavior)
We know that is a
-degree function with a positive leading coefficient. That is,
.
Since the degree of is even, its end behaviors match. And since the leading coefficient is positive, we know that both ends approach
as
goes in either direction, from which
So the first time
is going to be negative is when it intersects the
-axis at an
-intercept and it's going to dip below. This happens at
, which is the smallest intercept.
However, when it hits the next intercept, it's going to go back up again into positive territory, we know this happens at . And when it hits
, it's going to dip back into negative territory. Clearly, this is going to continue to snake around the intercepts until
.
To get the amount of integers below and/or on the -axis, we simply need to count the integers. For example, the amount of integers in between the
interval we got earlier, we subtract and add one.
integers, so there are four integers in this interval that produce a negative result.
Doing this with all of the other intervals, we have from Solution 2's result.
~quacker88
Solution 4 (Fast)
We know when an odd number of its factors are positive and negative. For example, to make the first factor positive,
. then there will be a even number of positive factors. We would do
to find all integers that work. In short we can generalize too:
But remember this only works when
because only then will there be a odd amount of positive and negative factors. So we can set
, for
Now we only have to solve:
~Wiselion
Video Solutions
https://youtu.be/3dfbWzOfJAI?t=4026
~ pi_is_3.14
~Education, The Study of Everything
~IceMatrix
https://www.youtube.com/watch?v=YDMMhSguq0w&list=PLeFyQ1uCoINM4D5Lgi5Y3KkfvQuYuIbj
-Walt S.
~savannahsolver
https://youtu.be/R220vbM_my8?t=463
~ amritvignesh0719062.0
See Also
2020 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.