Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2022 AMC 10B Problems/Problem 21"

(→‎Solution 5: Not very bashy)
(→‎Video Solutions)
 
(23 intermediate revisions by 8 users not shown)
Line 1: Line 1:
 +
{{duplicate|[[2022 AMC 10B Problems/Problem 21|2022 AMC 10B #21]] and [[2022 AMC 12B Problems/Problem 20|2022 AMC 12B #20]]}}
 +
 
==Problem==
 
==Problem==
  
Line 5: Line 7:
 
is <math>2x+1</math>. There is a unique polynomial of least degree with these two properties. What is the sum of
 
is <math>2x+1</math>. There is a unique polynomial of least degree with these two properties. What is the sum of
 
the squares of the coefficients of that polynomial?
 
the squares of the coefficients of that polynomial?
 +
 +
<math>\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 19 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 23</math>
  
 
==Solution 1 (Experimentation)==
 
==Solution 1 (Experimentation)==
Line 13: Line 17:
  
 
We are given:
 
We are given:
−
 
+
<cmath>111a + 12 = 101b + 21 = P(x).</cmath>
−
<cmath>111a + 12 = 101b + 21 = P(x)</cmath>.
 
−
 
 
 
We add <math>111</math> and <math>101</math> to each side and balance respectively:
 
We add <math>111</math> and <math>101</math> to each side and balance respectively:
−
 
+
<cmath>111(a - 1) + 123 = 101(b - 1) + 122 = P(x).</cmath>
−
<cmath>111(a - 1) + 123 = 101(b - 1) + 122 = P(x)</cmath>
+
We make the unit's digits equal:
−
 
+
<cmath>111(a - 1) + 123 = 101(b - 2) + 223 = P(x).</cmath>
−
We make the units digits equal:
 
−
 
 
−
<cmath>111(a - 1) + 123 = 101(b - 2) + 223 = P(x)</cmath>
 
−
 
 
 
We now notice that:
 
We now notice that:
 +
<cmath>111(a - 11) + 1233 = 101(b - 12) + 1233 = P(x).</cmath>
 +
Therefore <math>a = 11_{x} = x + 1</math>, <math>b = 12_{x} = x + 2</math>, and <math>P(x) = 1233_{x} = x^3 + 2x^2 + 3x + 3</math>. <math>3</math> is the minimal degree of <math>P(x)</math> since there is no way to influence the <math>x</math>‘s digit in <math>101b + 21</math> when <math>b</math> is an integer. The desired sum is <math>1^2 + 2^2 +3^2+ 3^2 = \boxed{\textbf{(E)} \ 23}</math>
  
−
<cmath>111(a - 11) + 1233 = 101(b - 12) + 1233 = P(x)</cmath>.
+
P.S. The four computational steps can be deduced through quick experimentation.
−
 
 
−
Therefore <math>a = 11_{x} = x + 1</math>, <math>b = 12_{x} = x + 2</math>, and <math>P(x) = 1233_{x} = x^3 + 2x^2 + 3x + 3</math>. <math>3</math> is the minimal degree of <math>P(x)</math> since there is no way to influence the <math>x</math>‘s digit in <math>101b + 21</math> when <math>b</math> is an integer. The desired sum is <math>1^2 + 2^2 + 3^2 + 3^2 = \boxed{\textbf{(E)} \ 23}</math>
 
−
 
 
−
P.S. The 4 computational steps can be deduced through quick experimentation.
 
  
 
~ numerophile
 
~ numerophile
Line 72: Line 68:
 
~qgcui
 
~qgcui
  
−
==Solution 4 (undetermined coefficients)==
+
==Solution 4 (Undetermined Coefficients)==
  
 
Notice that we cannot have the quotients equal to some constants, since the same constant will yield different constant terms for <math>P(x)</math> (which is bad) and different constants will yield different first coefficients (also bad). Thus, we try setting the quotients equal to linear terms (for minimizing degree).
 
Notice that we cannot have the quotients equal to some constants, since the same constant will yield different constant terms for <math>P(x)</math> (which is bad) and different constants will yield different first coefficients (also bad). Thus, we try setting the quotients equal to linear terms (for minimizing degree).
Line 82: Line 78:
 
~MathHayden
 
~MathHayden
  
−
 
+
==Solution 5 (Quick, but Not Quicker Than 2) ==
−
==Solution 5: Not very bashy ==
 
 
We construct the following equations in terms of <math>P(x)</math> and the information given by the problem:
 
We construct the following equations in terms of <math>P(x)</math> and the information given by the problem:
 
<cmath>\textbf{(1) } P(x)=(x^2+x+1)\cdot Q(x)+x+2</cmath>
 
<cmath>\textbf{(1) } P(x)=(x^2+x+1)\cdot Q(x)+x+2</cmath>
 
<cmath>\textbf{(2) } P(x)=(x^2+1)\cdot R(x)+2x+1</cmath>
 
<cmath>\textbf{(2) } P(x)=(x^2+1)\cdot R(x)+2x+1</cmath>
−
Upon inspection, <math>Q(x)</math> and <math>R(x)</math> cannot be constant, so the smallest possible degree of <math>P(x)</math> is <math>3,</math> and both <math>Q(x)</math> and <math>R(x)</math> are linear.
+
Upon inspection, <math>Q(x)</math> and <math>R(x)</math> cannot be constant, so the smallest possible degree of <math>P(x)</math> is <math>3,</math> and both <math>Q(x)</math> and <math>R(x)</math> are linear.  
  
−
Let <math>Q(x)=x-q</math> and <math>R(x)=x-r.</math>  
+
Let <math>Q(x)=x-q</math> and <math>R(x)=x-r.</math> We know there will be values for <math>q</math> and <math>r</math> that make the below equation hold, so we can assume that <math>P(x)</math> has a leading coefficient of <math>1</math>.
  
 
Substituting these values in, and setting <math>\textbf{(1)}</math> and <math>\textbf{(2)}</math> equal to each  other,
 
Substituting these values in, and setting <math>\textbf{(1)}</math> and <math>\textbf{(2)}</math> equal to each  other,
Line 101: Line 96:
 
<cmath>1^2+2^2+3^2+3^2=\boxed{\textbf{(E) }23}</cmath>
 
<cmath>1^2+2^2+3^2+3^2=\boxed{\textbf{(E) }23}</cmath>
  
−
==Video Solutions==
+
-Benedict T (countmath1)
 +
 
 +
==Solution 6 (Similar to Solution 3) ==
 +
By remainder theorem, the polynomial can be written as follows.
 +
 
 +
<cmath>P(x) = (x^2+x+1)Q_{1}(x)+x+2 = (x^2+1)Q_{2}(x)+2x+1</cmath>
 +
This is a timed exam, we can use the information given by answer choices. The answer choices tell us this is the polynomial with integer coefficients, and we need to find the polynomial with the least degree so we can assume both <math>Q_{1}(x)</math> and <math>Q_{2}(x)</math> are linear (the coefficient of x should be same).
 +
 
 +
Then we can write <math>P(x)</math> as a cubic polynomial.
 +
 
 +
<cmath>P(x) = (x^2+x+1)(ax+b)+x+2 = (x^2+1)(ax+c)+2x+1</cmath>
 +
Substituting <math>x=0,1,-1</math> to determine the value of <math>a</math> and <math>b</math>.
 +
 
 +
We have:
 +
<cmath>P(0) = b+2 = c+1</cmath>
 +
<cmath>P(1) = 3a+3b+3 = 2a+2c+3</cmath>
 +
<cmath>P(-1) = -a+b+1 = -2a+2c-1</cmath>
 +
 
 +
We can solve the simultaneous equations: <math>a=1,b=1,c=2</math>
 +
 
 +
Hence, <math>P(x)=(x^2+x+1)(x+1)+x+2=x^3+2x^2+3x+3</math>. The answer is <math>1^2+2^2+3^2+3^2=\boxed{\textbf{(E) }23}</math>
 +
 
 +
~PythZhou
 +
 
 +
 
 +
==Video Solution==
  
 
https://youtu.be/yGUur4vP_6k
 
https://youtu.be/yGUur4vP_6k
  
 
~ ThePuzzlr
 
~ ThePuzzlr
 +
 +
==Video Solution==
  
 
https://youtu.be/ELdhkqVyB9E
 
https://youtu.be/ELdhkqVyB9E
Line 111: Line 133:
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
  
−
== Video Solution by OmegaLearn using Circular Tangency ==
+
== Video Solution by OmegaLearn Using Polynomial Remainders ==
 
https://youtu.be/HdrbPiZHim0
 
https://youtu.be/HdrbPiZHim0
  
 
~ pi_is_3.14
 
~ pi_is_3.14
  
 +
==Video Solution by The Power of Logic(#20-#21)==
 +
 +
https://youtu.be/7FiTsDNMmgg
 +
==Video Solution by Interstigation==
 +
https://youtu.be/TvOHfDJ_2Ag
 +
 +
~Interstigation
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC10 box|year=2022|ab=B|num-b=20|num-a=22}}
 
{{AMC10 box|year=2022|ab=B|num-b=20|num-a=22}}
 +
{{AMC12 box|year=2022|ab=B|num-b=19|num-a=21}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 23:07, 19 September 2023

The following problem is from both the 2022 AMC 10B #21 and 2022 AMC 12B #20, so both problems redirect to this page.

Problem

Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1$, the remainder is $x+2$, and when $P(x)$ is divided by the polynomial $x^2+1$, the remainder is $2x+1$. There is a unique polynomial of least degree with these two properties. What is the sum of the squares of the coefficients of that polynomial?

$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 19 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 23$

Solution 1 (Experimentation)

Given that all the answer choices and coefficients are integers, we hope that $P(x)$ has positive integer coefficients.

Throughout this solution, we will express all polynomials in base $x$. E.g. $x^2 + x + 1 = 111_{x}$.

We are given: \[111a + 12 = 101b + 21 = P(x).\] We add $111$ and $101$ to each side and balance respectively: \[111(a - 1) + 123 = 101(b - 1) + 122 = P(x).\] We make the unit's digits equal: \[111(a - 1) + 123 = 101(b - 2) + 223 = P(x).\] We now notice that: \[111(a - 11) + 1233 = 101(b - 12) + 1233 = P(x).\] Therefore $a = 11_{x} = x + 1$, $b = 12_{x} = x + 2$, and $P(x) = 1233_{x} = x^3 + 2x^2 + 3x + 3$. $3$ is the minimal degree of $P(x)$ since there is no way to influence the $x$‘s digit in $101b + 21$ when $b$ is an integer. The desired sum is $1^2 + 2^2 +3^2+ 3^2 = \boxed{\textbf{(E)} \ 23}$

P.S. The four computational steps can be deduced through quick experimentation.

~ numerophile

Solution 2

Let $P(x) = Q(x)(x^2+x+1) + x + 2$, then $P(x) = Q(x)(x^2+1) + xQ(x) + x + 2$, therefore $xQ(x) + x + 2 \equiv 2x + 1 \pmod{x^2+1}$, or $xQ(x) \equiv x-1 \pmod{x^2+1}$. Clearly the minimum is when $Q(x) = x+1$, and expanding gives $P(x) = x^3+2x^2+3x+3$. Summing the squares of coefficients gives $\boxed{\textbf{(E)} \ 23}$

~mathfan2020

Solution 3

Let $P(x) = (x^2+x+1)Q_1(x) + x + 2$, then $P(x) = (x^2+1)Q_1(x) + xQ_1(x) + x + 2$

Also $P(x) = (x^2+1)Q_2(x) + 2x + 1$

We infer that $Q_1(x)$ and $Q_2(x)$ have same degree, we can assume $Q_1(x) = x + a$, and $Q_2(x) = x + b$, since $P(x)$ has least degree. If this cannot work, we will try quadratic, etc.

Then we get: $(x^2+1)(Q_1(x) - Q_2(x)) + xQ_1(x) - x + 1 = 0$

The constant term gives us: $(Q_1(x) - Q_2(x)) + 1 = 0$

So $Q_1(x) - Q_2(x) = -1$

Substituting this in gives: $-(x^2+1) + xQ_1(x) - x + 1 = 0$

Solving this equation, we get $Q_1(x) = x + 1$

Plugging this into our original equation we get $P(x) = x^3 + 2x^2 + 3x + 3$

Verify this works with $P(x) = (x^2+1)Q_2(x) + 2x + 1$

Therefore the answer is $1^2 + 2^2 + 3^2 + 3^2 = \boxed{\textbf{(E)} \ 23}$

~qgcui

Solution 4 (Undetermined Coefficients)

Notice that we cannot have the quotients equal to some constants, since the same constant will yield different constant terms for $P(x)$ (which is bad) and different constants will yield different first coefficients (also bad). Thus, we try setting the quotients equal to linear terms (for minimizing degree).

Let $P(x)=(x^2+x+1)(ax+b)+(x+2)$ and $P(x)=(x^2+1)(ax+c)+(2x+1)$. The quotients have the same $x$ coefficient, since $P(x)$ must have the same $x^3$ coefficient in both cases. Expanding, we get \[P(x)=ax^3+(a+b)x^2+(a+b+1)x+(b+2)\] and \[P(x)=ax^3+cx^2+(a+2)x+(c+1).\]

Equating coefficients, we get $b+2=c+1$, $a+b+1=a+2$, and $a+b=c$. From the second equation, we get $b=1$, then substituting into the first, $c=2$. Finally, from $a+b=c$, we have $a=1$. Now, $P(x)=(x^2+x+1)(ax+b)+(x+2)=(x^2+x+1)(x+1)+(x+2)=x^3+2x^2+3x+3$ and our answer is \[1^2+2^2+3^2+3^2=\boxed{\textbf{(E)} \ 23}.\]

~MathHayden

Solution 5 (Quick, but Not Quicker Than 2)

We construct the following equations in terms of $P(x)$ and the information given by the problem: \[\textbf{(1) } P(x)=(x^2+x+1)\cdot Q(x)+x+2\] \[\textbf{(2) } P(x)=(x^2+1)\cdot R(x)+2x+1\] Upon inspection, $Q(x)$ and $R(x)$ cannot be constant, so the smallest possible degree of $P(x)$ is $3,$ and both $Q(x)$ and $R(x)$ are linear.

Let $Q(x)=x-q$ and $R(x)=x-r.$ We know there will be values for $q$ and $r$ that make the below equation hold, so we can assume that $P(x)$ has a leading coefficient of $1$.

Substituting these values in, and setting $\textbf{(1)}$ and $\textbf{(2)}$ equal to each other, \[(x^2+x+1)(x-q)+x+2=(x^2+1)(x-r)+2x+1.\] We plug in $x=0$, yielding $r+1=q.$ Substituting this value into the above equation, \[(x^2+x+1)(x-r-1)+x+2=(x^2+1)(x-r)+2x+1.\] Letting $x=1,$ we conclude that $r=-2,$ so $R(x)=x+2.$ Therefore, \[P(x)=(x^2+1)(x+2)+2x+1 = x^3+2x^3+3x+3.\] The requested sum is \[1^2+2^2+3^2+3^2=\boxed{\textbf{(E) }23}\]

-Benedict T (countmath1)

Solution 6 (Similar to Solution 3)

By remainder theorem, the polynomial can be written as follows.

\[P(x) = (x^2+x+1)Q_{1}(x)+x+2 = (x^2+1)Q_{2}(x)+2x+1\] This is a timed exam, we can use the information given by answer choices. The answer choices tell us this is the polynomial with integer coefficients, and we need to find the polynomial with the least degree so we can assume both $Q_{1}(x)$ and $Q_{2}(x)$ are linear (the coefficient of x should be same).

Then we can write $P(x)$ as a cubic polynomial.

\[P(x) = (x^2+x+1)(ax+b)+x+2 = (x^2+1)(ax+c)+2x+1\] Substituting $x=0,1,-1$ to determine the value of $a$ and $b$.

We have: \[P(0) = b+2 = c+1\] \[P(1) = 3a+3b+3 = 2a+2c+3\] \[P(-1) = -a+b+1 = -2a+2c-1\]

We can solve the simultaneous equations: $a=1,b=1,c=2$

Hence, $P(x)=(x^2+x+1)(x+1)+x+2=x^3+2x^2+3x+3$. The answer is $1^2+2^2+3^2+3^2=\boxed{\textbf{(E) }23}$

~PythZhou


Video Solution

https://youtu.be/yGUur4vP_6k

~ ThePuzzlr

Video Solution

https://youtu.be/ELdhkqVyB9E

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution by OmegaLearn Using Polynomial Remainders

https://youtu.be/HdrbPiZHim0

~ pi_is_3.14

Video Solution by The Power of Logic(#20-#21)

https://youtu.be/7FiTsDNMmgg

Video Solution by Interstigation

https://youtu.be/TvOHfDJ_2Ag

~Interstigation

See Also

2022 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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
2022 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_10B_Problems/Problem_21&oldid=198423"