Difference between revisions of "2022 AMC 10B Problems/Problem 21"
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+ | {{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 | ||
− | |||
− | <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> | ||
− | + | P.S. The four computational steps can be deduced through quick experimentation. | |
− | |||
− | |||
− | |||
− | P.S. The | ||
~ numerophile | ~ numerophile | ||
Line 36: | Line 32: | ||
==Solution 2== | ==Solution 2== | ||
− | Let <math>P(x) = Q(x)(x^2+x+1) + x + 2</math>, then <math>P(x) = Q(x)(x^2+1) + xQ(x) + x + 2</math>, therefore <math>xQ(x) + x + 2 \equiv 2x + 1 \pmod{x^2+1}</math>, or <math>xQ(x) \equiv x-1 \pmod{x^2+1}</math>. Clearly the minimum is when <math>Q(x) = x | + | Let <math>P(x) = Q(x)(x^2+x+1) + x + 2</math>, then <math>P(x) = Q(x)(x^2+1) + xQ(x) + x + 2</math>, therefore <math>xQ(x) + x + 2 \equiv 2x + 1 \pmod{x^2+1}</math>, or <math>xQ(x) \equiv x-1 \pmod{x^2+1}</math>. Clearly the minimum is when <math>Q(x) = x+1</math>, and expanding gives <math>P(x) = x^3+2x^2+3x+3</math>. Summing the squares of coefficients gives <math>\boxed{\textbf{(E)} \ 23}</math> |
~mathfan2020 | ~mathfan2020 | ||
Line 42: | Line 38: | ||
==Solution 3== | ==Solution 3== | ||
− | Let <math>P(x) = | + | Let <math>P(x) = (x^2+x+1)Q_1(x) + x + 2</math>, |
− | then <math>P(x) = | + | then <math>P(x) = (x^2+1)Q_1(x) + xQ_1(x) + x + 2</math> |
− | Also <math>P(x) = | + | |
+ | Also <math>P(x) = (x^2+1)Q_2(x) + 2x + 1</math> | ||
+ | |||
+ | We infer that <math>Q_1(x)</math> and <math>Q_2(x)</math> have same degree, we can assume <math>Q_1(x) = x + a </math>, and <math>Q_2(x) = x + b </math>, since <math>P(x)</math> has least degree. If this cannot work, we will try quadratic, etc. | ||
Then we get: | Then we get: | ||
− | <math>( Q_1(x) - Q_2(x) | + | <math>(x^2+1)(Q_1(x) - Q_2(x)) + xQ_1(x) - x + 1 = 0 </math> |
The constant term gives us: | The constant term gives us: | ||
− | <math>( Q_1(x) - Q_2(x)) + 1 = 0 </math> | + | <math>(Q_1(x) - Q_2(x)) + 1 = 0 </math> |
− | So | + | So <math>Q_1(x) - Q_2(x) = -1 </math> |
− | <math> | ||
Substituting this in gives: | Substituting this in gives: | ||
<math>-(x^2+1) + xQ_1(x) - x + 1 = 0</math> | <math>-(x^2+1) + xQ_1(x) - x + 1 = 0</math> | ||
− | <math>Q_1(x) = x + 1 </math> | + | Solving this equation, we get |
+ | <math>Q_1(x) = x + 1 </math> | ||
+ | |||
+ | Plugging this into our original equation we get | ||
+ | <math>P(x) = x^3 + 2x^2 + 3x + 3 </math> | ||
− | <math> P(x) = x^ | + | Verify this works with <math>P(x) = (x^2+1)Q_2(x) + 2x + 1</math> |
+ | |||
+ | Therefore the answer is <math>1^2 + 2^2 + 3^2 + 3^2 = \boxed{\textbf{(E)} \ 23}</math> | ||
~qgcui | ~qgcui | ||
− | ==Video | + | ==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). | ||
+ | |||
+ | Let <math>P(x)=(x^2+x+1)(ax+b)+(x+2)</math> and <math>P(x)=(x^2+1)(ax+c)+(2x+1)</math>. The quotients have the same <math>x</math> coefficient, since <math>P(x)</math> must have the same <math>x^3</math> coefficient in both cases. Expanding, we get <cmath>P(x)=ax^3+(a+b)x^2+(a+b+1)x+(b+2)</cmath> and <cmath>P(x)=ax^3+cx^2+(a+2)x+(c+1).</cmath> | ||
+ | |||
+ | Equating coefficients, we get <math>b+2=c+1</math>, <math>a+b+1=a+2</math>, and <math>a+b=c</math>. From the second equation, we get <math>b=1</math>, then substituting into the first, <math>c=2</math>. Finally, from <math>a+b=c</math>, we have <math>a=1</math>. Now, <math>P(x)=(x^2+x+1)(ax+b)+(x+2)=(x^2+x+1)(x+1)+(x+2)=x^3+2x^2+3x+3</math> and our answer is <cmath>1^2+2^2+3^2+3^2=\boxed{\textbf{(E)} \ 23}.</cmath> | ||
+ | |||
+ | ~MathHayden | ||
+ | |||
+ | ==Solution 5 (Quick, but Not Quicker Than Solution 2) == | ||
+ | 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{(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. | ||
+ | |||
+ | 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, | ||
+ | <cmath>(x^2+x+1)(x-q)+x+2=(x^2+1)(x-r)+2x+1.</cmath> | ||
+ | We plug in <math>x=0</math>, yielding <math>r+1=q.</math> Substituting this value into the above equation, | ||
+ | <cmath>(x^2+x+1)(x-r-1)+x+2=(x^2+1)(x-r)+2x+1.</cmath> | ||
+ | Letting <math>x=1,</math> we conclude that <math>r=-2,</math> so <math>R(x)=x+2.</math> | ||
+ | Therefore, | ||
+ | <cmath>P(x)=(x^2+1)(x+2)+2x+1 = x^3+2x^3+3x+3.</cmath> | ||
+ | The requested sum is | ||
+ | <cmath>1^2+2^2+3^2+3^2=\boxed{\textbf{(E) }23}</cmath> | ||
+ | |||
+ | -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 | ||
+ | |||
+ | ==Solution 7 (Bounding with Answer Choices)== | ||
+ | We are given: | ||
+ | |||
+ | <math>P(x) \equiv x + 2 \pmod{x^2+x+1}</math> | ||
+ | |||
+ | <math>P(x) \equiv 2x + 1 \pmod{x^2+1}</math> | ||
+ | |||
+ | Since the problem asks us about the coefficients of the polynomial, we should evaluate <math>P(x)</math> at <math>1</math>. | ||
+ | |||
+ | Our congruences now become: | ||
+ | |||
+ | <math>P(1) \equiv 1 + 2 \pmod{(1)^2+(1)+1}</math> <math>\implies P(1) \equiv 3 \pmod{3} \implies P(1) \equiv 0 \pmod{3}</math>. | ||
+ | |||
+ | <math>P(1) \equiv 2(1) + 1 \pmod{(1)^2+1}</math> <math>\implies P(1) \equiv 3 \pmod{2} \implies P(1) \equiv 1 \pmod{2}</math>. | ||
+ | |||
+ | We now know that the coefficients of the polynomial sum to an odd multiple of 3. | ||
+ | |||
+ | We now begin bounding the value of the sum of the coefficients squared. For our answer, the sum of the coefficients squared must be between <math>10</math> and <math>23</math>. Trying a quadratic polynomial, if <math>P(1) = 3</math>, then the maximum of the coefficients squared is <math>9</math> (<math>3^2+0^2+0^2</math> is the maximum, any other permutation can be trivially shown to be lower). If <math>P(1) = 9</math>, the minimum of the coefficients squared is <math>27</math> (<math>3^2+3^2+3^2</math>, once again, any other permutation can be trivially shown to be higher). Any higher value of <math>P(1)</math> obviously will not work, so <math>P(x)</math> cannot be a quadratic. | ||
+ | |||
+ | Since a quadratic did not work, we now move up to a cubic. We start with <math>P(1) = 3</math>. Similarly to the quadratic, there is no way to get an answer within the range of the answer choices. | ||
+ | |||
+ | However, moving up to <math>P(1) = 9</math>, we get that the minimum of the coefficients squared is <math>81/4</math> (which we get from all 4 coefficients being equal at <math>9/4</math>). Since this is the minimum, any other answers will be higher. We get that <math>\boxed{\textbf{(E) }23}</math> is our answer, since it is the only answer choice greater than <math>81/4</math>. | ||
+ | |||
+ | -SwordOfJustice | ||
+ | |||
+ | ==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 74: | Line 157: | ||
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
− | == Video Solution by OmegaLearn | + | == 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 | ||
+ | |||
+ | ==Video Solution by TheBeautyofMath== | ||
+ | https://youtu.be/UEQOPjlqh94 | ||
+ | ~IceMatrix | ||
== 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 23:04, 29 June 2024
- The following problem is from both the 2022 AMC 10B #21 and 2022 AMC 12B #20, so both problems redirect to this page.
Contents
- 1 Problem
- 2 Solution 1 (Experimentation)
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4 (Undetermined Coefficients)
- 6 Solution 5 (Quick, but Not Quicker Than Solution 2)
- 7 Solution 6 (Similar to Solution 3)
- 8 Solution 7 (Bounding with Answer Choices)
- 9 Video Solution
- 10 Video Solution
- 11 Video Solution by OmegaLearn Using Polynomial Remainders
- 12 Video Solution by The Power of Logic(#20-#21)
- 13 Video Solution by Interstigation
- 14 Video Solution by TheBeautyofMath
- 15 See Also
Problem
Let be a polynomial with rational coefficients such that when
is divided by the polynomial
, the remainder is
, and when
is divided by the polynomial
, the remainder
is
. 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?
Solution 1 (Experimentation)
Given that all the answer choices and coefficients are integers, we hope that has positive integer coefficients.
Throughout this solution, we will express all polynomials in base . E.g.
.
We are given:
We add
and
to each side and balance respectively:
We make the unit's digits equal:
We now notice that:
Therefore
,
, and
.
is the minimal degree of
since there is no way to influence the
‘s digit in
when
is an integer. The desired sum is
P.S. The four computational steps can be deduced through quick experimentation.
~ numerophile
Solution 2
Let , then
, therefore
, or
. Clearly the minimum is when
, and expanding gives
. Summing the squares of coefficients gives
~mathfan2020
Solution 3
Let ,
then
Also
We infer that and
have same degree, we can assume
, and
, since
has least degree. If this cannot work, we will try quadratic, etc.
Then we get:
The constant term gives us:
So
Substituting this in gives:
Solving this equation, we get
Plugging this into our original equation we get
Verify this works with
Therefore the answer is
~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 (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 and
. The quotients have the same
coefficient, since
must have the same
coefficient in both cases. Expanding, we get
and
Equating coefficients, we get ,
, and
. From the second equation, we get
, then substituting into the first,
. Finally, from
, we have
. Now,
and our answer is
~MathHayden
Solution 5 (Quick, but Not Quicker Than Solution 2)
We construct the following equations in terms of and the information given by the problem:
Upon inspection,
and
cannot be constant, so the smallest possible degree of
is
and both
and
are linear.
Let and
We know there will be values for
and
that make the below equation hold, so we can assume that
has a leading coefficient of
.
Substituting these values in, and setting and
equal to each other,
We plug in
, yielding
Substituting this value into the above equation,
Letting
we conclude that
so
Therefore,
The requested sum is
-Benedict T (countmath1)
Solution 6 (Similar to Solution 3)
By remainder theorem, the polynomial can be written as follows.
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
and
are linear (the coefficient of x should be same).
Then we can write as a cubic polynomial.
Substituting
to determine the value of
and
.
We have:
We can solve the simultaneous equations:
Hence, . The answer is
~PythZhou
Solution 7 (Bounding with Answer Choices)
We are given:
Since the problem asks us about the coefficients of the polynomial, we should evaluate at
.
Our congruences now become:
.
.
We now know that the coefficients of the polynomial sum to an odd multiple of 3.
We now begin bounding the value of the sum of the coefficients squared. For our answer, the sum of the coefficients squared must be between and
. Trying a quadratic polynomial, if
, then the maximum of the coefficients squared is
(
is the maximum, any other permutation can be trivially shown to be lower). If
, the minimum of the coefficients squared is
(
, once again, any other permutation can be trivially shown to be higher). Any higher value of
obviously will not work, so
cannot be a quadratic.
Since a quadratic did not work, we now move up to a cubic. We start with . Similarly to the quadratic, there is no way to get an answer within the range of the answer choices.
However, moving up to , we get that the minimum of the coefficients squared is
(which we get from all 4 coefficients being equal at
). Since this is the minimum, any other answers will be higher. We get that
is our answer, since it is the only answer choice greater than
.
-SwordOfJustice
Video Solution
~ ThePuzzlr
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by OmegaLearn Using Polynomial Remainders
~ pi_is_3.14
Video Solution by The Power of Logic(#20-#21)
Video Solution by Interstigation
~Interstigation
Video Solution by TheBeautyofMath
~IceMatrix
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.